Chapter IV
Limits of Functions of a Positive Integral Variable

50. Functions of a positive integral variable. In Chapter II we discussed the notion of a function of a real variable x, and illustrated the discussion by a large number of examples of such functions. And the reader will remember that there was one important particular with regard to which the functions which we took as illustrations differed very widely. Some were defined for all values of x, some for rational values only, some for integral values only, and so on.

Consider, for example, the following functions: (i) x, (ii) x, (iii) the denominator of x, (iv) the square root of the product of the numerator and the denominator of x, (v) the largest prime factor of x, (vi) the product of x and the largest prime factor of x, (vii) the xth prime number, (viii) the height measured in inches of convict x in Dartmoor prison.

Then the aggregates of values of x for which these functions are defined or, as we may say, the fields of definition of the functions, consist of (i) all values of x, (ii) all positive values of x, (iii) all rational values of x, (iv) all positive rational values of x, (v) all integral values of x, (vi), (vii) all positive integral values of x, (viii) a certain number of positive integral values of x, viz., 12, …, N, where N is the total number of convicts at Dartmoor at a given moment of time.36

Now let us consider a function, such as (vii) above, which is defined for all positive integral values of x and no others. This function may be regarded from two slightly different points of view. We may consider it, as has so far been our custom, as a function of the real variable x defined for some only of the values of x, viz. positive integral values, and say that for all other values of x the definition fails. Or we may leave values of x other than positive integral values entirely out of account, and regard our function as a function of the positive integral variable n, whose values are the positive integers

1,2,3,4,.
In this case we may write
y = φ(n)
and regard y now as a function of n defined for all values of n.

It is obvious that any function of x defined for all values of x gives rise to a function of n defined for all values of n. Thus from the function y = x2 we deduce the function y = n2 by merely omitting from consideration all values of x other than positive integers, and the corresponding values of y. On the other hand from any function of n we can deduce any number of functions of x by merely assigning values to y, corresponding to values of x other than positive integral values, in any way we please.

51. Interpolation. The problem of determining a function of x which shall assume, for all positive integral values of x, values agreeing with those of a given function of n, is of extreme importance in higher mathematics. It is called the problem of functional interpolation.

Were the problem however merely that of finding some function of x to fulfil the condition stated, it would of course present no difficulty whatever. We could, as explained above, simply fill in the missing values as we pleased: we might indeed simply regard the given values of the function of n as all the values of the function of x and say that the definition of the latter function failed for all other values of x. But such purely theoretical solutions are obviously not what is usually wanted. What is usually wanted is some formula involving x (of as simple a kind as possible) which assumes the given values for x = 1, 2, ….

In some cases, especially when the function of n is itself defined by a formula, there is an obvious solution. If for example y = φ(n), where φ(n) is a function of n, such as n2 or cosnπ, which would have a meaning even were n not a positive integer, we naturally take our function of x to be y = φ(x). But even in this very simple case it is easy to write down other almost equally obvious solutions of the problem. For example

y = φ(x) + sinxπ
assumes the value φ(n) for x = n, since sinnπ = 0.

In other cases φ(n) may be defined by a formula, such as (1)n, which ceases to define for some values of x (as here in the case of fractional values of x with even denominators, or irrational values). But it may be possible to transform the formula in such a way that it does define for all values of x. In this case, for example,

(1)n = cosnπ,
if n is an integer, and the problem of interpolation is solved by the function cosxπ.

In other cases φ(x) may be defined for some values of x other than positive integers, but not for all. Thus from y = nn we are led to y = xx. This expression has a meaning for some only of the remaining values of x. If for simplicity we confine ourselves to positive values of x, then xx has a meaning for all rational values of x, in virtue of the definitions of fractional powers adopted in elementary algebra. But when x is irrational xx has (so far as we are in a position to say at the present moment) no meaning at all. Thus in this case the problem of interpolation at once leads us to consider the question of extending our definitions in such a way that xx shall have a meaning even when x is irrational. We shall see later on how the desired extension may be effected.

Again, consider the case in which

y = 1 2n = n!.
In this case there is no obvious formula in x which reduces to n! for x = n, as x! means nothing for values of x other than the positive integers. This is a case in which attempts to solve the problem of interpolation have led to important advances in mathematics. For mathematicians have succeeded in discovering a function (the Gamma-function) which possesses the desired property and many other interesting and important properties besides.

52. Finite and infinite classes. Before we proceed further it is necessary to make a few remarks about certain ideas of an abstract and logical nature which are of constant occurrence in Pure Mathematics.

In the first place, the reader is probably familiar with the notion of a class. It is unnecessary to discuss here any logical difficulties which may be involved in the notion of a ‘class’: roughly speaking we may say that a class is the aggregate or collection of all the entities or objects which possess a certain property, simple or complex. Thus we have the class of British subjects, or members of Parliament, or positive integers, or real numbers.

Moreover, the reader has probably an idea of what is meant by a finite or infinite class. Thus the class of British subjects is a finite class: the aggregate of all British subjects, past, present, and future, has a finite number n, though of course we cannot tell at present the actual value of n. The class of present British subjects, on the other hand, has a number n which could be ascertained by counting, were the methods of the census effective enough.

On the other hand the class of positive integers is not finite but infinite. This may be expressed more precisely as follows. If n is any positive integer, such as 1000, 1, 000, 000 or any number we like to think of, then there are more than n positive integers. Thus, if the number we think of is 1, 000, 000, there are obviously at least 1, 000, 001 positive integers. Similarly the class of rational numbers, or of real numbers, is infinite. It is convenient to express this by saying that there are an infinite number of positive integers, or rational numbers, or real numbers. But the reader must be careful always to remember that by saying this we mean simply that the class in question has not a finite number of members such as 1000 or 1, 000, 000.

53. Properties possessed by a function of n for large values of n. We may now return to the ‘functions of n’ which we were discussing in §§ 5051. They have many points of difference from the functions of x which we discussed in Chap. II. But there is one fundamental characteristic which the two classes of functions have in common: the values of the variable for which they are defined form an infinite class. It is this fact which forms the basis of all the considerations which follow and which, as we shall see in the next chapter, apply, mutatis mutandis, to functions of x as well.

Suppose that φ(n) is any function of n, and that P is any property which φ(n) may or may not have, such as that of being a positive integer or of being greater than 1. Consider, for each of the values n = 1, 2, 3, …, whether φ(n) has the property P or not. Then there are three possibilities:—

(a) φ(n) may have the property P for all values of n, or for all values of n except a finite number N of such values:

(b) φ(n) may have the property for no values of n, or only for a finite number N of such values:

(c) neither (a) nor (b) may be true.

If (b) is true, the values of n for which φ(n) has the property form a finite class. If (a) is true, the values of n for which φ(n) has not the property form a finite class. In the third case neither class is finite. Let us consider some particular cases.

(1) Let φ(n) = n, and let P be the property of being a positive integer. Then φ(n) has the property P for all values of n.

If on the other hand P denotes the property of being a positive integer greater than or equal to 1000, then φ(n) has the property for all values of n except a finite number of values of n, viz. 12, 3, …, 999. In either of these cases (a) is true.

(2) If φ(n) = n, and P is the property of being less than 1000, then (b) is true.

(3) If φ(n) = n, and P is the property of being odd, then (c) is true. For φ(n) is odd if n is odd and even if n is even, and both the odd and the even values of n form an infinite class.

Example. Consider, in each of the following cases, whether (a), (b), or (c) is true:

(i) φ(n) = n, P being the property of being a perfect square,

(ii) φ(n) = pn, where pn denotes the nth prime number, P being the
property of being odd,

(iii) φ(n) = pn, P being the property of being even,

(iv) φ(n) = pn, P being the property φ(n) > n,

(v) φ(n) = 1 (1)n(1/n), P being the property φ(n) < 1,

(vi) φ(n) = 1 (1)n(1/n), P being the property φ(n) < 2,

(vii) φ(n) = 1000{1 + (1)n}/n, P being the property φ(n) < 1,

(viii) φ(n) = 1/n, P being the property φ(n) < .001,

(ix) φ(n) = (1)n/n, P being the property φ(n)< .001,

(x) φ(n) = 10,000/n, or (1)n10,000/n, P being either of the properties φ(n) < .001 or φ(n)< .001,

(xi) φ(n) = (n 1)/(n + 1), P being the property 1 φ(n) < .0001.

54. Let us now suppose that φ(n) and P are such that the assertion (a) is true, i.e. that φ(n) has the property P, if not for all values of n, at any rate for all values of n except a finite number N of such values. We may denote these exceptional values by

n1,n2,,nN.
There is of course no reason why these N values should be the first N values 12, …, N, though, as the preceding examples show, this is frequently the case in practice. But whether this is so or not we know that φ(n) has the property P if n > nN. Thus the nth prime is odd if n > 2, n = 2 being the only exception to the statement; and 1/n < .001 if n > 1000, the first 1000 values of n being the exceptions; and
1000{1 + (1)n}/n < 1
if n > 2000, the exceptional values being 246, …, 2000. That is to say, in each of these cases the property is possessed for all values of n from a definite value onwards.

We shall frequently express this by saying that φ(n) has the property for large, or very large, or all sufficiently large values of n. Thus when we say that φ(n) has the property P (which will as a rule be a property expressed by some relation of inequality) for large values of n, what we mean is that we can determine some definite number, n0 say, such that φ(n) has the property for all values of n greater than or equal to n0. This number n0, in the examples considered above, may be taken to be any number greater than nN, the greatest of the exceptional numbers: it is most natural to take it to be nN + 1.

Thus we may say that ‘all large primes are odd’, or that ‘1/n is less than .001 for large values of n’. And the reader must make himself familiar with the use of the word large in statements of this kind. Large is in fact a word which, standing by itself, has no more absolute meaning in mathematics than in the language of common life. It is a truism that in common life a number which is large in one connection is small in another; 6 goals is a large score in a football match, but 6 runs is not a large score in a cricket match; and 400 runs is a large score, but £400 is not a large income: and so of course in mathematics large generally means large enough, and what is large enough for one purpose may not be large enough for another.

We know now what is meant by the assertion ‘φ(n) has the property P for large values of n’. It is with assertions of this kind that we shall be concerned throughout this chapter.

55. The phrase ‘n tends to infinity’. There is a somewhat different way of looking at the matter which it is natural to adopt. Suppose that n assumes successively the values 12, 3, …. The word ‘successively’ naturally suggests succession in time, and we may suppose n, if we like, to assume these values at successive moments of time (e.g. at the beginnings of successive seconds). Then as the seconds pass n gets larger and larger and there is no limit to the extent of its increase. However large a number we may think of (e.g. 2, 147, 483, 647), a time will come when n has become larger than this number.

It is convenient to have a short phrase to express this unending growth of n, and we shall say that n tends to infinity, or n , this last symbol being usually employed as an abbreviation for ‘infinity’. The phrase ‘tends to’ like the word ‘successively’ naturally suggests the idea of change in time, and it is convenient to think of the variation of n as accomplished in time in the manner described above. This however is a mere matter of convenience. The variable n is a purely logical entity which has in itself nothing to do with time.

The reader cannot too strongly impress upon himself that when we say that n ‘tends to ’ we mean simply that n is supposed to assume a series of values which increase continually and without limit. There is no number ‘infinity’: such an equation as

n =
is as it stands absolutely meaningless: n cannot be equal to , because ‘equal to ’ means nothing. So far in fact the symbol  means nothing at all except in the one phrase ‘tends to ’, the meaning of which we have explained above. Later on we shall learn how to attach a meaning to other phrases involving the symbol , but the reader will always have to bear in mind

(1) that  by itself means nothing, although phrases containing it sometimes mean something,

(2) that in every case in which a phrase containing the symbol  means something it will do so simply because we have previously attached a meaning to this particular phrase by means of a special definition.

Now it is clear that if φ(n) has the property P for large values of n, and if n ‘tends to ’, in the sense which we have just explained, then n will ultimately assume values large enough to ensure that φ(n) has the property P. And so another way of putting the question ‘what properties has φ(n) for sufficiently large values of n?’ is ‘how does φ(n) behave as n tends to ?’

56. The behaviour of a function of n as n tends to infinity. We shall now proceed, in the light of the remarks made in the preceding sections, to consider the meaning of some kinds of statements which are perpetually occurring in higher mathematics. Let us consider, for example, the two following statements: (a1/n is small for large values of n, (b1 (1/n) is nearly equal to 1 for large values of n. Obvious as they may seem, there is a good deal in them which will repay the reader’s attention. Let us take (a) first, as being slightly the simpler.

We have already considered the statement ‘1/n is less than .01 for large values of n’. This, we saw, means that the inequality 1/n < .01 is true for all values of n greater than some definite value, in fact greater than 100. Similarly it is true that ‘1/n is less than .0001 for large values of n’: in fact 1/n < .0001 if n > 10, 000. And instead of .01 or .0001 we might take .000001 or .00000001, or indeed any positive number we like.

It is obviously convenient to have some way of expressing the fact that any such statement as ‘1/n is less than .01 for large values of n’ is true, when we substitute for .01 any smaller number, such as .0001 or .000001 or any other number we care to choose. And clearly we can do this by saying that ‘however small ε may be (provided of course it is positive), then 1/n < ε for sufficiently large values of n’. That this is true is obvious. For 1/n < ε if n > 1/ε, so that our ‘sufficiently large’ values of n need only all be greater than 1/ε. The assertion is however a complex one, in that it really stands for the whole class of assertions which we obtain by giving to ε special values such as .01. And of course the smaller ε is, and the larger 1/ε, the larger must be the least of the ‘sufficiently large’ values of n: values which are sufficiently large when ε has one value are inadequate when it has a smaller.

The last statement italicised is what is really meant by the statement (a), that 1/n is small when n is large. Similarly (b) really means “if φ(n) = 1 (1/n), then the statement ‘1 φ(n) < ε for sufficiently large values of n’ is true whatever positive value suchas.01or.0001 we attribute to ε”. That the statement (b) is true is obvious from the fact that 1 φ(n) = 1/n.

There is another way in which it is common to state the facts expressed by the assertions (a) and (b). This is suggested at once by § 55. Instead of saying ‘1/n is small for large values of n’ we say ‘1/n tends to 0 as n tends to ’. Similarly we say that ‘1 (1/n) tends to 1 as n tends to ’: and these statements are to be regarded as precisely equivalent to (a) and (b). Thus the statements

 ‘1/n is small when n is large’,  ‘1/n tends to 0 as n tends to ’,

are equivalent to one another and to the more formal statement

‘if ε is any positive number, however small, then 1/n < ε for sufficiently large values of n’,

or to the still more formal statement

‘if ε is any positive number, however small, then we can find a number n0 such that 1/n < ε for all values of n greater than or equal to n0’.

The number n0 which occurs in the last statement is of course a function of ε. We shall sometimes emphasize this fact by writing n0 in the form n0(ε).

The reader should imagine himself confronted by an opponent who questions the truth of the statement. He would name a series of numbers growing smaller and smaller. He might begin with .001. The reader would reply that 1/n < .001 as soon as n > 1000. The opponent would be bound to admit this, but would try again with some smaller number, such as .0000001. The reader would reply that 1/n < .0000001 as soon as n > 10,000,000: and so on. In this simple case it is evident that the reader would always have the better of the argument.

We shall now introduce yet another way of expressing this property of the function 1/n. We shall say that ‘the limit of 1/n as n tends to  is 0’, a statement which we may express symbolically in the form

lim n1 n = 0,
or simply lim(1/n) = 0. We shall also sometimes write ‘1/n 0 as n ’, which may be read ‘1/n tends to 0 as n tends to ’; or simply ‘1/n 0’. In the same way we shall write
lim n1 1 n = 1, lim 1 1 n = 1,
or 1 (1/n) 1.

57. Now let us consider a different example: let φ(n) = n2. Then ‘n2 is large when n is large’. This statement is equivalent to the more formal statements

‘if Δ is any positive number, however large, then n2 > Δ for sufficiently large values of n’,

‘we can find a number n0(Δ) such that n2 > Δ for all values of n greater than or equal to n0(Δ)’.

And it is natural in this case to say that ‘n2 tends to  as n tends to ’, or ‘n2 tends to  with n’, and to write

n2 .

Finally consider the function φ(n) = n2. In this case φ(n) is large, but negative, when n is large, and we naturally say that ‘ n2 tends to  as n tends to ’ and write

n2 .
And the use of the symbol  in this sense suggests that it will sometimes be convenient to write n2 + for n2 and generally to use  + instead of , in order to secure greater uniformity of notation.

But we must once more repeat that in all these statements the symbols + mean nothing whatever by themselves, and only acquire a meaning when they occur in certain special connections in virtue of the explanations which we have just given.

58. Definition of a limit. After the discussion which precedes the reader should be in a position to appreciate the general notion of a limit. Roughly we may say that φ(n) tends to a limit l as n tends to if φ(n) is nearly equal to l when n is large. But although the meaning of this statement should be clear enough after the preceding explanations, it is not, as it stands, precise enough to serve as a strict mathematical definition. It is, in fact, equivalent to a whole class of statements of the type ‘for sufficiently large values of n, φ(n) differs from l by less than ε’. This statement has to be true for ε = .01 or .0001 or any positive number; and for any such value of ε it has to be true for any value of n after a certain definite value n0(ε), though the smaller ε is the larger, as a rule, will be this value n0(ε).

We accordingly frame the following formal definition:

DEFINITION I. The function φ(n) is said to tend to the limit l as n tends to , if, however small be the positive number ε, φ(n) differs from l by less than ε for sufficiently large values of n; that is to say if, however small be the positive number ε, we can determine a number n0(ε) corresponding to ε, such that φ(n) differs from l by less than ε for all values of n greater than or equal to n0(ε).

It is usual to denote the difference between φ(n) and l, taken positively, by φ(n) l. It is equal to φ(n) l or to l φ(n), whichever is positive, and agrees with the definition of the modulus of φ(n) l, as given in Chap. III, though at present we are only considering real values, positive or negative.

With this notation the definition may be stated more shortly as follows: ‘if, given any positive number, ε, however small, we can find n0(ε) so that φ(n) l < ε when n n0(ε), then we say that φ(n) tends to the limit l as n tends to , and write

lim nφ(n) = l ’.

Sometimes we may omit the ‘n ’; and sometimes it is convenient, for brevity, to write φ(n) l.

The reader will find it instructive to work out, in a few simple cases, the explicit expression of n0 as a function of ε. Thus if φ(n) = 1/n then l = 0, and the condition reduces to 1/n < ε for n n0, which is satisfied if n0 = 1 + [1/ε].37 There is one and only one case in which the same n0 will do for all values of ε. If, from a certain value N of n onwards, φ(n) is constant, say equal to C, then it is evident that φ(n) C = 0 for n N, so that the inequality φ(n) C< ε is satisfied for n N and all positive values of ε. And if φ(n) l< ε for n N and all positive values of ε, then it is evident that φ(n) = l when n N, so that φ(n) is constant for all such values of n.

59. The definition of a limit may be illustrated geometrically as follows. The graph of φ(n) consists of a number of points corresponding to the values n = 1, 23, ….

Draw the line y = l, and the parallel lines y = l ε, y = l + ε at distance ε from it. Then

lim nφ(n) = l,


pict

Fig. 27.

if, when once these lines have been drawn, no matter how close they may be together, we can always draw a line x = n0, as in the figure, in such a way that the point of the graph on this line, and all points to the right of it, lie between them. We shall find this geometrical way of looking at our definition particularly useful when we come to deal with functions defined for all values of a real variable and not merely for positive integral values.

60. So much for functions of n which tend to a limit as n tends to . We must now frame corresponding definitions for functions which, like the functions n2 or  n2, tend to positive or negative infinity. The reader should by now find no difficulty in appreciating the point of

DEFINITION II. The function φ(n) is said to tend to  + (positive infinity) with n, if, when any number Δ, however large, is assigned, we can determine n0(Δ) so that φ(n) > Δ when n n0(Δ); that is to say if, however large Δ may be, φ(n) > Δ for sufficiently large values of n.

Another, less precise, form of statement is ‘if we can make φ(n) as large as we please by sufficiently increasing n’. This is open to the objection that it obscures a fundamental point, viz. that φ(n) must be greater than Δ for all values of n such that n n0(Δ), and not merely for some such values. But there is no harm in using this form of expression if we are clear what it means.

When φ(n) tends to  + we write

φ(n) +.
We may leave it to the reader to frame the corresponding definition for functions which tend to negative infinity.

61. Some points concerning the definitions. The reader should be careful to observe the following points.

(1) We may obviously alter the values of φ(n) for any finite number of values of n, in any way we please, without in the least affecting the behaviour of φ(n) as n tends to . For example 1/n tends to 0 as n tends to . We may deduce any number of new functions from 1/n by altering a finite number of its values. For instance we may consider the function φ(n) which is equal to 3 for n = 12, 7, 11, 101, 107, 109237 and equal to 1/n for all other values of n. For this function, just as for the original function 1/n, lim φ(n) = 0. Similarly, for the function φ(n) which is equal to 3 if n = 12, 7, 11, 101, 107, 109237, and to n2 otherwise, it is true that φ(n) +.

(2) On the other hand we cannot as a rule alter an infinite number of the values of φ(n) without affecting fundamentally its behaviour as n tends to . If for example we altered the function 1/n by changing its value to 1 whenever n is a multiple of 100, it would no longer be true that lim φ(n) = 0. So long as a finite number of values only were affected we could always choose the number n0 of the definition so as to be greater than the greatest of the values of n for which φ(n) was altered. In the examples above, for instance, we could always take n0 > 237, and indeed we should be compelled to do so as soon as our imaginary opponent of § 56 had assigned a value of ε as small as 3 (in the first example) or a value of Δ as great as 3 (in the second). But now however large n0 may be there will be greater values of n for which φ(n) has been altered.

(3) In applying the test of Definition I it is of course absolutely essential that we should have φ(n) l < ε not merely when n = n0 but when n n0, i.e. for n0 and for all larger values of n. It is obvious, for example, that, if φ(n) is the function last considered, then given ε we can choose n0 so that φ(n) < ε when n = n0: we have only to choose a sufficiently large value of n which is not a multiple of 100. But, when n0 is thus chosen, it is not true that φ(n) < ε when n n0: all the multiples of 100 which are greater than n0 are exceptions to this statement.

(4) If φ(n) is always greater than l, we can replace φ(n) l by φ(n) l. Thus the test whether 1/n tends to the limit 0 as n tends to  is simply whether 1/n < ε when n n0. If however φ(n) = (1)n/n, then l is again 0, but φ(n) l is sometimes positive and sometimes negative. In such a case we must state the condition in the form φ(n) l < ε, for example, in this particular case, in the form φ(n) < ε.

(5) The limit l may itself be one of the actual values of φ(n). Thus if φ(n) = 0 for all values of n, it is obvious that lim φ(n) = 0. Again, if we had, in (2) and (3) above, altered the value of the function, when n is a multiple of 100, to 0 instead of to 1, we should have obtained a function φ(n) which is equal to 0 when n is a multiple of 100 and to 1/n otherwise. The limit of this function as n tends to  is still obviously zero. This limit is itself the value of the function for an infinite number of values of n, viz. all multiples of 100.

On the other hand the limit itself need not andingeneralwillnot be the value of the function for any value of n. This is sufficiently obvious in the case of φ(n) = 1/n. The limit is zero; but the function is never equal to zero for any value of n.

The reader cannot impress these facts too strongly on his mind. A limit is not a value of the function: it is something quite distinct from these values, though it is defined by its relations to them and may possibly be equal to some of them. For the functions

φ(n) = 0,1,
the limit is equal to all the values of φ(n): for
φ(n) = 1/n,(1)n/n,1 + (1/n),1 + {(1)n/n}
it is not equal to any value of φ(n): for
φ(n) = (sin 1 2nπ)/n,1 + {(sin 1 2nπ)/n}
(whose limits as n tends to  are easily seen to be 0 and 1, since sin 1 2nπ is never numerically greater than 1) the limit is equal to the value which φ(n) assumes for all even values of n, but the values assumed for odd values of n are all different from the limit and from one another.

(6) A function may be always numerically very large when n is very large without tending either to  + or to  . A sufficient illustration of this is given by φ(n) = (1)nn. A function can only tend to  + or to  if, after a certain value of n, it maintains a constant sign.

Examples XXIII. Consider the behaviour of the following functions of n as n tends to :

1. φ(n) = nk, where k is a positive or negative integer or rational fraction. If k is positive, then nk tends to  + with n. If k is negative, then limnk = 0. If k = 0, then nk = 1 for all values of n. Hence limnk = 1.

The reader will find it instructive, even in so simple a case as this, to write down a formal proof that the conditions of our definitions are satisfied. Take for instance the case of k > 0. Let Δ be any assigned number, however large. We wish to choose n0 so that nk > Δ when n n0. We have in fact only to take for n0 any number greater than Δk. If e.g. k = 4, then n4 > 10,000 when n 11, n4 > 100,000,000 when n 101, and so on.

2. φ(n) = pn, where pn is the nth prime number. If there were only a finite number of primes then φ(n) would be defined only for a finite number of values of n. There are however, as was first shown by Euclid, infinitely many primes. Euclid’s proof is as follows. If there are only a finite number of primes, let them be 12, 3, 5, 7, 11, … N. Consider the number 1 + (1 2 3 5 7 11N). This number is evidently not divisible by any of 23, 5, … N, since the remainder when it is divided by any of these numbers is 1. It is therefore not divisible by any prime save 1, and is therefore itself prime, which is contrary to our hypothesis.

It is moreover obvious that φ(n) > n for all values of n (save n = 1, 23). Hence φ(n) +.

3. Let φ(n) be the number of primes less than n. Here again φ(n) +.

4. φ(n) = [αn], where α is any positive number. Here

φ(n) = 0(0 n < 1/α),φ(n) = 1(1/α n < 2/α),
and so on; and φ(n) +.

5. If φ(n) = 1,000,000/n, then limφ(n) = 0: and if ψ(n) = n/1,000,000, then ψ(n) +. These conclusions are in no way affected by the fact that at first φ(n) is much larger than ψ(n), being in fact larger until n = 1,000,000.

6. φ(n) = 1/{n (1)n}, n (1)n, n{1 (1)n}. The first function tends to 0, the second to  + , the third does not tend either to a limit or to  + .

7. φ(n) = (sinnθπ)/n, where θ is any real number. Here φ(n)< 1/n, since sinnθπ1, and limφ(n) = 0.

8. φ(n) = (sinnθπ)/n, (acos2nθ + bsin2nθ)/n, where a and b are any real numbers.

9. φ(n) = sinnθπ. If θ is integral then φ(n) = 0 for all values of n, and therefore limφ(n) = 0.

Next let θ be rational, e.g. θ = p/q, where p and q are positive integers. Let n = aq + b where a is the quotient and b the remainder when n is divided by q. Then sin(npπ/q) = (1)ap sin(bpπ/q). Suppose, for example, p even; then, as n increases from 0 to q 1, φ(n) takes the values

0,sin(pπ/q),sin(2pπ/q),sin{(q 1)pπ/q}.
When n increases from q to 2q 1 these values are repeated; and so also as n goes from 2q to 3q 1, 3q to 4q 1, and so on. Thus the values of φ(n) form a perpetual cyclic repetition of a finite series of different values. It is evident that when this is the case φ(n) cannot tend to a limit, nor to  + , nor to  , as n tends to infinity.

The case in which θ is irrational is a little more difficult. It is discussed in the next set of examples.

62. Oscillating Functions.

DEFINITION. When φ(n) does not tend to a limit, nor to  + , nor to  , as n tends to , we say that φ(n) oscillates as n tends to .

A function φ(n) certainly oscillates if its values form, as in the case considered in the last example above, a continual repetition of a cycle of values. But of course it may oscillate without possessing this peculiarity. Oscillation is defined in a purely negative manner: a function oscillates when it does not do certain other things.

The simplest example of an oscillatory function is given by

φ(n) = (1)n,
which is equal to  + 1 when n is even and to  1 when n is odd. In this case the values recur cyclically. But consider
φ(n) = (1)n + (1/n),
the values of which are
1 + 1,1 + (1/2), 1 + (1/3),1 + (1/4), 1 + (1/5),.
When n is large every value is nearly equal to  + 1 or  1, and obviously φ(n) does not tend to a limit or to  + or to  , and therefore it oscillates: but the values do not recur. It is to be observed that in this case every value of φ(n) is numerically less than or equal to 3/2. Similarly
φ(n) = (1)n100 + (1000/n)
oscillates. When n is large, every value is nearly equal to 100 or to  100. The numerically greatest value is 900 (for n = 1). But now consider φ(n) = (1)nn, the values of which are 1, 2, 3, 4, 5, …. This function oscillates, for it does not tend to a limit, nor to  + , nor to  . And in this case we cannot assign any limit beyond which the numerical value of the terms does not rise. The distinction between these two examples suggests a further definition.

DEFINITION. If φ(n) oscillates as n tends to , then φ(n) will be said to oscillate finitely or infinitely according as it is or is not possible to assign a number K such that all the values of φ(n) are numerically less than K, i.e. φ(n) < K for all values of n.

These definitions, as well as those of §§ 58 and 60, are further illustrated in the following examples.

Examples XXIV. Consider the behaviour as n tends to of the following functions:

1. (1)n, 5 + 3(1)n, (1,000,000/n) + (1)n, 1,000,000(1)n + (1/n).

2. (1)nn, 1,000,000 + (1)nn.

3. 1,000,000 n, (1)n(1,000,000 n).

4. n{1 + (1)n}. In this case the values of φ(n) are

0,4,0,8,0,12,0,16,.
The odd terms are all zero and the even terms tend to  + : φ(n) oscillates infinitely.

5. n2 + (1)n2n. The second term oscillates infinitely, but the first is very much larger than the second when n is large. In fact φ(n) n2 2n and n2 2n = (n 1)2 1 is greater than any assigned value Δ if n > 1 + Δ + 1. Thus φ(n) +. It should be observed that in this case φ(2k + 1) is always less than φ(2k), so that the function progresses to infinity by a continual series of steps forwards and backwards. It does not however ‘oscillate’ according to our definition of the term.

6. n2{1 + (1)n}, (1)nn2 + n, n3 + (1)nn2.

7. sinnθπ. We have already seen (Exs. XXIII. 9) that φ(n) oscillates finitely when θ is rational, unless θ is an integer, when φ(n) = 0, φ(n) 0.

The case in which θ is irrational is a little more difficult. But it is not difficult to see that φ(n) still oscillates finitely. We can without loss of generality suppose 0 < θ < 1. In the first place φ(n)< 1. Hence φ(n) must oscillate finitely or tend to a limit. We shall consider whether the second alternative is really possible. Let us suppose that

limsinnθπ = l.
Then, however small ε may be, we can choose n0 so that sinnθπ lies between l ε and l + ε for all values of n greater than or equal to n0. Hence sin(n + 1)θπ sinnθπ is numerically less than 2ε for all such values of n, and so sin 1 2θπcos(n + 1 2)θπ< ε.

Hence

cos(n + 1 2)θπ = cosnθπcos 1 2θπ sinnθπsin 1 2θπ
must be numerically less than ε/sin 1 2θπ. Similarly
cos(n 1 2)θπ = cosnθπcos 1 2θπ + sinnθπsin 1 2θπ
must be numerically less than ε/sin 1 2θπ; and so each of cosnθπcos 1 2θπ, sinnθπsin 1 2θπ must be numerically less than ε/sin 1 2θπ. That is to say, cosnθπcos 1 2θπ is very small if n is large, and this can only be the case if cosnθπ is very small. Similarly sinnθπ must be very small, so that l must be zero. But it is impossible that cosnθπ and sinnθπ can both be very small, as the sum of their squares is unity. Thus the hypothesis that sinnθπ tends to a limit l is impossible, and therefore sinnθπ oscillates as n tends to .

The reader should consider with particular care the argument ‘cosnθπcos 1 2θπ is very small, and this can only be the case if cosnθπ is very small’. Why, he may ask, should it not be the other factor cos 1 2θπ which is ‘very small’? The answer is to be found, of course, in the meaning of the phrase ‘very small’ as used in this connection. When we say ‘φ(n) is very small’ for large values of n, we mean that we can choose n0 so that φ(n) is numerically smaller than any assigned number, if n n0. Such an assertion is palpably absurd when made of a fixed number such as cos 1 2θπ, which is not zero.

Prove similarly that cosnθπ oscillates finitely, unless θ is an even integer.

8. sinnθπ + (1/n), sinnθπ + 1, sinnθπ + n, (1)n sinnθπ.

9. acosnθπ + bsinnθπ, sin2nθπ, acos2nθπ + bsin2nθπ.

10. a + bn + (1)n(c + dn) + ecosnθπ + fsinnθπ.

11. nsinnθπ. If θ is integral, then φ(n) = 0, φ(n) 0. If θ is rational but not integral, or irrational, then φ(n) oscillates infinitely.

12. n(acos2nθπ + bsin2nθπ). In this case φ(n) tends to  + if a and b are both positive, but to  if both are negative. Consider the special cases in which a = 0, b > 0, or a > 0, b = 0, or a = 0, b = 0. If a and b have opposite signs φ(n) generally oscillates infinitely. Consider any exceptional cases.

13. sin(n2θπ). If θ is integral, then φ(n) 0. Otherwise φ(n) oscillates finitely, as may be shown by arguments similar to though more complex than those used in Exs. XXIII. 9 and XXIV. 7.38

14. sin(n!θπ). If θ has a rational value p/q, then n!θ is certainly integral for all values of n greater than or equal to q. Hence φ(n) 0. The case in which θ is irrational cannot be dealt with without the aid of considerations of a much more difficult character.

15. cos(n!θπ), acos2(n!θπ) + bsin2(n!θπ), where θ is rational.

16. an [bn], (1)n(an [bn]).

17. [n], (1)n[n], n [n].

18. The smallest prime factor of n. When n is a prime, φ(n) = n. When n is even, φ(n) = 2. Thus φ(n) oscillates infinitely.

19. The largest prime factor of n.

20. The number of days in the year n A.D.

Examples XXV. 1. If φ(n) + and ψ(n) φ(n) for all values of n, then ψ(n) +.

2. If φ(n) 0, and ψ(n)φ(n)for all values of n, then ψ(n) 0.

3. If limφ(n)= 0, then limφ(n) = 0.

4. If φ(n) tends to a limit or oscillates finitely, and ψ(n)φ(n) when n n0, then ψ(n) tends to a limit or oscillates finitely.

5. If φ(n) tends to  + , or to  , or oscillates infinitely, and

ψ(n)φ(n)
when n n0, then ψ(n) tends to  + or to  or oscillates infinitely.

6. ‘If φ(n) oscillates and, however great be n0, we can find values of n greater than n0 for which ψ(n) > φ(n), and values of n greater than n0 for which ψ(n) < φ(n), then ψ(n) oscillates’. Is this true? If not give an example to the contrary.

7. If φ(n) l as n , then also φ(n + p) l, p being any fixed integer. [This follows at once from the definition. Similarly we see that if φ(n) tends to  + or  or oscillates so also does φ(n + p).]

8. The same conclusions hold (except in the case of oscillation) if p varies with n but is always numerically less than a fixed positive integer N; or if p varies with n in any way, so long as it is always positive.

9. Determine the least value of n0 for which it is true that

(a)n2 + 2n > 999,999(n n 0),(b) n2 + 2n > 1,000,000(n n 0).

10. Determine the least value of n0 for which it is true that

(a)n + (1)n > 1000(n n 0),(b) n + (1)n > 1,000,000(n n 0).

11. Determine the least value of n0 for which it is true that

(a)n2 + 2n > Δ(n n 0),(b) n + (1)n > Δ(n n 0),
Δ being any positive number.

[(an0 = [Δ + 1]: (bn0 = 1 + [Δ] or 2 + [Δ], according as [Δ] is odd or even, i.e. n0 = 1 + [Δ] + 1 2{1 + (1)[Δ]}.]

12. Determine the least value of n0 such that

(a)n/(n2 + 1) < .0001,(b) (1/n) + {(1)n/n2}< .00001,
when n n0. [Let us take the latter case. In the first place
(1/n) + {(1)n/n2}(n + 1)/n2,
and it is easy to see that the least value of n0, such that (n + 1)/n2 < .000001 when n n0, is 1,000,002. But the inequality given is satisfied by n = 1,000,001, and this is the value of n0 required.]

63. Some general theorems with regard to limits. A. The behaviour of the sum of two functions whose behaviour is known.

THEOREM I. If φ(n) and ψ(n) tend to limits ab, then φ(n) + ψ(n) tends to the limit a + b.

This is almost obvious.39 The argument which the reader will at once form in his mind is roughly this: ‘when n is large, φ(n) is nearly equal to a and ψ(n) to b, and therefore their sum is nearly equal to a + b’. It is well to state the argument quite formally, however.

Let ε be any assigned positive number (e.g. .001, .0000001, …). We require to show that a number n0 can be found such that

φ(n) + ψ(n) a b < ε, (1)

when n n0. Now by a proposition proved in Chap. III (more generally indeed than we need here) the modulus of the sum of two numbers is less than or equal to the sum of their moduli. Thus

φ(n) + ψ(n) a bφ(n) a + ψ(n) b.
It follows that the desired condition will certainly be satisfied if n0 can be so chosen that

φ(n) a + ψ(n) b < ε, (2)

when n n0. But this is certainly the case. For since lim φ(n) = a we can, by the definition of a limit, find n1 so that φ(n) a < ε when n n1, and this however small ε may be. Nothing prevents our taking ε = 1 2ε, so that φ(n) a < 1 2ε when n n1. Similarly we can find n2 so that ψ(n) b < 1 2ε when n n2. Now take n0 to be the greater of the two numbers n1n2. Then φ(n) a < 1 2ε and ψ(n) b < 1 2ε when n n0, and therefore (2) is satisfied and the theorem is proved.

The argument may be concisely stated thus: since limφ(n) = a and limψ(n) = b, we can choose n1n2 so that

φ(n) a< 1 2ε(n n1),ψ(n) b< 1 2ε(n n2);
and then, if n is not less than either n1 or n2,
φ(n) + ψ(n) a bφ(n) a+ ψ(n) b< ε;
and therefore
lim{φ(n) + ψ(n)}= a + b.

64. Results subsidiary to Theorem I. The reader should have no difficulty in verifying the following subsidiary results.

1. If φ(n) tends to a limit, but ψ(n) tends to  + or to  or oscillates finitely or infinitely, then φ(n) + ψ(n) behaves like ψ(n).

2. If φ(n) +, and ψ(n) + or oscillates finitely, then φ(n) + ψ(n) +.

In this statement we may obviously change + into  throughout.

3. If φ(n) + and ψ(n) , then φ(n) + ψ(n) may tend either to a limit or to  + or to  or may oscillate either finitely or infinitely.

These five possibilities are illustrated in order by (i) φ(n) = n, ψ(n) = n, (ii) φ(n) = n2, ψ(n) = n, (iii) φ(n) = n, ψ(n) = n2, (iv) φ(n) = n + (1)n, ψ(n) = n, (v) φ(n) = n2 + (1)nn, ψ(n) = n2. The reader should construct additional examples of each case.

4. If φ(n) + and ψ(n) oscillates infinitely, then φ(n) + ψ(n) may tend to  + or oscillate infinitely, but cannot tend to a limit, or to  , or oscillate finitely.

For ψ(n) = {φ(n) + ψ(n)}φ(n); and, if φ(n) + ψ(n) behaved in any of the three last ways, it would follow, from the previous results, that ψ(n) , which is not the case. As examples of the two cases which are possible, consider (i) φ(n) = n2, ψ(n) = (1)nn, (ii) φ(n) = n, ψ(n) = (1)nn2. Here again the signs of  + and  may be permuted throughout.

5. If φ(n) and ψ(n) both oscillate finitely, then φ(n) + ψ(n) must tend to a limit or oscillate finitely.

As examples take

(i) φ(n) = (1)n,ψ(n) = (1)n+1,(ii) φ(n) = ψ(n) = (1)n.

6. If φ(n) oscillates finitely, and ψ(n) infinitely, then φ(n) + ψ(n) oscillates infinitely.

For φ(n) is in absolute value always less than a certain constant, say K. On the other hand ψ(n), since it oscillates infinitely, must assume values numerically greater than any assignable number (e.g. 10K, 100K, …). Hence φ(n) + ψ(n) must assume values numerically greater than any assignable number (e.g. 9K, 99K, …). Hence φ(n) + ψ(n) must either tend to  + or  or oscillate infinitely. But if it tended to  + then

ψ(n) = {φ(n) + ψ(n)}φ(n)
would also tend to  + , in virtue of the preceding results. Thus φ(n) + ψ(n) cannot tend to  + , nor, for similar reasons, to  : hence it oscillates infinitely.

7. If both φ(n) and ψ(n) oscillate infinitely, then φ(n) + ψ(n) may tend to a limit, or to  + , or to  , or oscillate either finitely or infinitely.

Suppose, for instance, that φ(n) = (1)nn, while ψ(n) is in turn each of the functions (1)n+1n, {1 + (1)n+1}n, {1 + (1)n}n, (1)n+1(n + 1), (1)nn. We thus obtain examples of all five possibilities.

The results 1–7 cover all the cases which are really distinct. Before passing on to consider the product of two functions, we may point out that the result of Theorem I may be immediately extended to the sum of three or more functions which tend to limits as n .

65. B. The behaviour of the product of two functions whose behaviour is known. We can now prove a similar set of theorems concerning the product of two functions. The principal result is the following.

THEOREM II. If lim φ(n) = a and lim ψ(n) = b, then

lim φ(n)ψ(n) = ab.

Let

φ(n) = a + φ1(n),ψ(n) = b + ψ1(n),
so that lim φ1(n) = 0 and lim ψ1(n) = 0. Then
φ(n)ψ(n) = ab + aψ1(n) + bφ1(n) + φ1(n)ψ1(n).
Hence the numerical value of the difference φ(n)ψ(n) ab is certainly not greater than the sum of the numerical values of aψ1(n), bφ1(n), φ1(n)ψ1(n). From this it follows that
lim{φ(n)ψ(n) ab} = 0,
which proves the theorem.

The following is a strictly formal proof. We have

φ(n)ψ(n) abaψ1(n)+ bφ1(n)+ φ1(n)ψ1(n).
Assuming that neither a nor b is zero, we may choose n0 so that
φ1(n)< 1 3ε/b,ψ1(n)< 1 3ε/a,
when n n0. Then
φ(n)ψ(n) ab< 1 3ε + 1 3ε + {1 9ε2/(ab)},
which is certainly less than ε if ε < 1 3ab. That is to say we can choose n0 so that φ(n)ψ(n) ab< ε when n n0, and so the theorem follows. The reader should supply a proof for the case in which at least one of a and b is zero.

We need hardly point out that this theorem, like Theorem I, may be immediately extended to the product of any number of functions of n. There is also a series of subsidiary theorems concerning products analogous to those stated in § 64 for sums. We must distinguish now six different ways in which φ(n) may behave as n tends to . It may (1) tend to a limit other than zero, (2) tend to zero, (3a) tend to  + , (3b) tend to  , (4) oscillate finitely, (5) oscillate infinitely. It is not necessary, as a rule, to take account separately of (3a) and (3b), as the results for one case may be deduced from those for the other by a change of sign.

To state these subsidiary theorems at length would occupy more space than we can afford. We select the two which follow as examples, leaving the verification of them to the reader. He will find it an instructive exercise to formulate some of the remaining theorems himself.

(i) If φ(n) + and ψ(n) oscillates finitely, then φ(n)ψ(n) must tend to  + or to  or oscillate infinitely.

Examples of these three possibilities may be obtained by taking φ(n) to be n and ψ(n) to be one of the three functions 2 + (1)n, 2 (1)n, (1)n.

(ii) If φ(n) and ψ(n) oscillate finitely, then φ(n)ψ(n) must tend to a limit whichmaybezero or oscillate finitely.

For examples, take (aφ(n) = ψ(n) = (1)n, (bφ(n) = 1 + (1)n, ψ(n) = 1 (1)n, and (cφ(n) = cos 1 3nπ, ψ(n) = sin 1 3nπ.

A particular case of Theorem II which is important is that in which ψ(n) is constant. The theorem then asserts simply that lim kφ(n) = ka if lim φ(n) = a. To this we may join the subsidiary theorem that if φ(n) + then kφ(n) + or kφ(n) , according as k is positive or negative, unless k = 0, when of course kφ(n) = 0 for all values of n and lim kφ(n) = 0. And if φ(n) oscillates finitely or infinitely, then so does kφ(n), unless k = 0.

66. C. The behaviour of the difference or quotient of two functions whose behaviour is known. There is, of course, a similar set of theorems for the difference of two given functions, which are obvious corollaries from what precedes. In order to deal with the quotient

φ(n) ψ(n),
we begin with the following theorem.

THEOREM III. If lim φ(n) = a, and a is not zero, then

lim 1 φ(n) = 1 a.

Let

φ(n) = a + φ1(n),
so that lim φ1(n) = 0. Then
1 φ(n) 1 a = φ1(n) aa + φ1(n),
and it is plain, since lim φ1(n) = 0, that we can choose n0 so that this is smaller than any assigned number ε when n n0.

From Theorems II and III we can at once deduce the principal theorem for quotients, viz. 

THEOREM IV. If lim φ(n) = a and lim ψ(n) = b, and b is not zero, then

lim φ(n) ψ(n) = a b.

The reader will again find it instructive to formulate, prove, and illustrate by examples some of the ‘subsidiary theorems’ corresponding to Theorems III and IV.

67.

THEOREM V. If R{φ(n),ψ(n),χ(n),}is any rational function of φ(n), ψ(n), χ(n), …, i.e. any function of the form

P{φ(n),ψ(n),χ(n),}/Q{φ(n),ψ(n),χ(n),},
where P and Q denote polynomials in φ(n), ψ(n), χ(n), …: and if
lim φ(n) = a, lim ψ(n) = b, lim χ(n) = c,,
and
Q(a,b,c,)0;
then
lim R{φ(n),ψ(n),χ(n),} = R(a,b,c,).

For P is a sum of a finite number of terms of the type

A{φ(n)}p{ψ(n)}q,
where A is a constant and pq, … positive integers. This term, by Theorem II (or rather by its obvious extension to the product of any number of functions) tends to the limit Aapbq, and so P tends to the limit P(a,b,c,), by the similar extension of Theorem I. Similarly Q tends to Q(a,b,c,); and the result then follows from Theorem IV.

68. The preceding general theorem may be applied to the following very important particular problem: what is the behaviour of the most general rational function of n, viz.

S(n) = a0np + a 1np1 + + a p b0nq + b1nq1 + + bq ,
as n tends to ?40

In order to apply the theorem we transform S(n) by writing it in the form

npq (a 0 + a1 n + + ap np)(b0 + b1 n + + bq nq) .
The function in curly brackets is of the form R{φ(n)}, where φ(n) = 1/n, and therefore tends, as n tends to , to the limit R(0) = a0/b0. Now npq 0 if p < q; npq = 1 and npq 1 if p = q; and npq + if p > q. Hence, by Theorem II,

lim S(n) = 0(p < q), lim S(n) = a0/b0(p = q), S(n) +(p > q, a0/b0 positive), S(n) (p > q, a0/b0 negative).

Examples XXVI. 1. What is the behaviour of the functions

n 1 n + 12,(1)n n 1 n + 12,n2 + 1 n ,(1)nn2 + 1 n ,
as n ?

2. Which (if any) of the functions

1/(cos21 2nπ + nsin21 2nπ),1/{n(cos21 2nπ + nsin21 2nπ)}, (ncos21 2nπ + sin21 2nπ)/{n(cos21 2nπ + nsin21 2nπ)}

tend to a limit as n ?

3. Denoting by S(n) the general rational function of n considered above, show that in all cases

lim S(n + 1) S(n) = 1,lim S{n + (1/n)} S(n) = 1.

69. Functions of n which increase steadily with n. A special but particularly important class of functions of n is formed by those whose variation as n tends to  is always in the same direction, that is to say those which always increase (or always decrease) as n increases. Since φ(n) always increases if φ(n) always decreases, it is not necessary to consider the two kinds of functions separately; for theorems proved for one kind can at once be extended to the other.

DEFINITION. The function φ(n) will be said to increase steadily with n if φ(n + 1) φ(n) for all values of n.

It is to be observed that we do not exclude the case in which φ(n) has the same value for several values of n; all we exclude is possible decrease. Thus the function

φ(n) = 2n + (1)n,
whose values for n = 0, 1, 2, 3, 4, … are
1,1,5,5,9,9,
is said to increase steadily with n. Our definition would indeed include even functions which remain constant from some value of n onwards; thus φ(n) = 1 steadily increases according to our definition. However, as these functions are extremely special ones, and as there can be no doubt as to their behaviour as n tends to , this apparent incongruity in the definition is not a serious defect.

There is one exceedingly important theorem concerning functions of this class.

THEOREM. If φ(n) steadily increases with n, then either (i) φ(n) tends to a limit as n tends to , or (ii) φ(n) +.

That is to say, while there are in general five alternatives as to the behaviour of a function, there are two only for this special kind of function.

This theorem is a simple corollary of Dedekind’s Theorem (§ 17). We divide the real numbers ξ into two classes L and R, putting ξ in L or R according as φ(n) ξ for some value of n (and so of course for all greater values), or φ(n) < ξ for all values of n.

The class L certainly exists; the class R may or may not. If it does not, then, given any number Δ, however large, φ(n) > Δ for all sufficiently large values of n, and so

φ(n) +.

If on the other hand R exists, the classes L and R form a section of the real numbers in the sense of § 17. Let a be the number corresponding to the section, and let ε be any positive number. Then φ(n) < a + ε for all values of n, and so, since ε is arbitrary, φ(n) a. On the other hand φ(n) > a ε for some value of n, and so for all sufficiently large values. Thus

a ε < φ(n) a
for all sufficiently large values of n; i.e. 
φ(n) a.

It should be observed that in general φ(n) < a for all values of n; for if φ(n) is equal to a for any value of n it must be equal to a for all greater values of n. Thus φ(n) can never be equal to a except in the case in which the values of φ(n) are ultimately all the same. If this is so, a is the largest member of L; otherwise L has no largest member.

COR 1. If φ(n) increases steadily with n, then it will tend to a limit or to  + according as it is or is not possible to find a number K such that φ(n) < K for all values of n.

We shall find this corollary exceedingly useful later on.

COR 2. If φ(n) increases steadily with n, and φ(n) < K for all values of n, then φ(n) tends to a limit and this limit is less than or equal to K.

It should be noticed that the limit may be equal to K: if e.g. φ(n) = 3 (1/n), then every value of φ(n) is less than 3, but the limit is equal to 3.

COR 3. If φ(n) increases steadily with n, and tends to a limit, then

φ(n) lim φ(n)
for all values of n.

The reader should write out for himself the corresponding theorems and corollaries for the case in which φ(n) decreases as n increases.

70. The great importance of these theorems lies in the fact that they give us (what we have so far been without) a means of deciding, in a great many cases, whether a given function of n does or does not tend to a limit as n , without requiring us to be able to guess or otherwise infer beforehand what the limit is. If we know what the limit, if there is one, must be, we can use the test

φ(n) l < ε(n n0) :
as for example in the case of φ(n) = 1/n, where it is obvious that the limit can only be zero. But suppose we have to determine whether
φ(n) = 1 + 1 nn
tends to a limit. In this case it is not obvious what the limit, if there is one, will be: and it is evident that the test above, which involves l, cannot be used, at any rate directly, to decide whether l exists or not.

Of course the test can sometimes be used indirectly, to prove by means of a reductio ad absurdum that l cannot exist. If e.g. φ(n) = (1)n, it is clear that l would have to be equal to 1 and also equal to  1, which is obviously impossible.

71. Alternative proof of Weierstrass’s Theorem of §19. The results of §69 enable us to give an alternative proof of the important theorem proved in §19.

If we divide PQ into two equal parts, one at least of them must contain infinitely many points of S. We select the one which does, or, if both do, we select the left-hand half; and we denote the selected half by P1Q1 (Fig. 28). If P1Q1 is the left-hand half, P1 is the same point as P.


pict

Fig. 28.

Similarly, if we divide P1Q1 into two halves, one at least of them must contain infinitely many points of S. We select the half P2Q2 which does so, or, if both do so, we select the left-hand half. Proceeding in this way we can define a sequence of intervals

PQ,P1Q1,P2Q2,P3Q3,,
each of which is a half of its predecessor, and each of which contains infinitely many points of S.

The points P, P1, P2, … progress steadily from left to right, and so Pn tends to a limiting position T. Similarly Qn tends to a limiting position T. But TT is plainly less than PnQn, whatever the value of n; and PnQn, being equal to PQ/2n, tends to zero. Hence T coincides with T, and Pn and Qn both tend to T.

Then T is a point of accumulation of S. For suppose that ξ is its coordinate, and consider any interval of the type [ξ ε,ξ + ε]. If n is sufficiently large, PnQn will lie entirely inside this interval.41 Hence [ξ ε,ξ + ε] contains infinitely many points of S.

72. The limit of xn as n tends to . Let us apply the results of § 69 to the particularly important case in which φ(n) = xn. If x = 1 then φ(n) = 1, lim φ(n) = 1, and if x = 0 then φ(n) = 0, lim φ(n) = 0, so that these special cases need not detain us.

First, suppose x positive. Then, since φ(n + 1) = xφ(n), φ(n) increases with n if x > 1, decreases as n increases if x < 1.

If x > 1, then xn must tend either to a limit (which must obviously be greater than 1) or to  + . Suppose it tends to a limit l. Then lim φ(n + 1) = lim φ(n) = l, by Exs. XXV. 7; but

lim φ(n + 1) = lim xφ(n) = x lim φ(n) = xl,
and therefore l = xl: and as x and l are both greater than 1, this is impossible. Hence
xn +(x > 1).

Example. The reader may give an alternative proof, showing by the binomial theorem that xn > 1 + nδ if δ is positive and x = 1 + δ, and so that

xn +.

On the other hand xn is a decreasing function if x < 1, and must therefore tend to a limit or to  . Since xn is positive the second alternative may be ignored. Thus lim xn = l, say, and as above l = xl, so that l must be zero. Hence

lim xn = 0(0 < x < 1).

Example. Prove as in the preceding example that (1/x)n tends to  + if 0 < x < 1, and deduce that xn tends to 0.

We have finally to consider the case in which x is negative. If 1 < x < 0 and x = y, so that 0 < y < 1, then it follows from what precedes that lim yn = 0 and therefore lim xn = 0. If x = 1 it is obvious that xn oscillates, taking the values 11 alternatively. Finally if x < 1, and x = y, so that y > 1, then yn tends to  + , and therefore xn takes values, both positive and negative, numerically greater than any assigned number. Hence xn oscillates infinitely. To sum up:

φ(n) = xn + (x > 1), lim φ(n) = 1 (x = 1), lim φ(n) = 0 (1 < x < 1),  φ(n) oscillates finitely (x = 1),  φ(n) oscillates infinitely (x < 1).

Examples XXVII.42 1. If φ(n) is positive and φ(n + 1) > Kφ(n), where K > 1, for all values of n, then φ(n) +.

[For

φ(n) > Kφ(n 1) > K2φ(n 2) > Kn1φ(1),
from which the conclusion follows at once, as Kn .]

2. The same result is true if the conditions above stated are satisfied only when n n0.

3. If φ(n) is positive and φ(n + 1) < Kφ(n), where 0 < K < 1, then limφ(n) = 0. This result also is true if the conditions are satisfied only when n n0.

4. If φ(n + 1)< Kφ(n) when n n0, and 0 < K < 1, then limφ(n) = 0.

5. If φ(n) is positive and lim{φ(n + 1)}/{φ(n)}= l > 1, then φ(n) +.

[For we can determine n0 so that {φ(n + 1)}/{φ(n)}> K > 1 when n n0: we may, e.g., take K halfway between 1 and l. Now apply Ex. 1.]

6. If lim{φ(n + 1)}/{φ(n)}= l, where l is numerically less than unity, then limφ(n) = 0. [This follows from Ex. 4 as Ex. 5 follows from Ex. 1.]

7. Determine the behaviour, as n , of φ(n) = nrxn, where r is any positive integer.

[If x = 0 then φ(n) = 0 for all values of n, and φ(n) 0. In all other cases

φ(n + 1) φ(n) = n + 1 n rx x.
First suppose x positive. Then φ(n) +if x > 1 (Ex. 5) and φ(n) 0 if x < 1 (Ex. 6). If x = 1, then φ(n) = nr +. Next suppose x negative. Then φ(n)= nrxn tends to  + if x1 and to 0 if x< 1. Hence φ(n) oscillates infinitely if x 1 and φ(n) 0 if 1 < x < 0.]

8. Discuss nrxn in the same way. [The results are the same, except that φ(n) 0 when x = 1 or  1.]

9. Draw up a table to show how nkxn behaves as n , for all real values of x, and all positive and negative integral values of k.

[The reader will observe that the value of k is immaterial except in the special cases when x = 1 or  1. Since lim{(n + 1)/n}k = 1, whether k be positive or negative, the limit of the ratio φ(n + 1)/φ(n) depends only on x, and the behaviour of φ(n) is in general dominated by the factor xn. The factor nk only asserts itself when x is numerically equal to 1.]

10. Prove that if x is positive then xn 1 as n . [Suppose, e.g., x > 1. Then xx, x3, … is a decreasing sequence, and xn > 1 for all values of n. Thus xn l, where l 1. But if l > 1 we can find values of n, as large as we please, for which xn > l or x > ln; and, since ln + as n , this is impossible.]

11. nn 1. [For n + 1n + 1 < nn if (n + 1)n < nn+1 or {1 + (1/n)}n < n, which is certainly satisfied if n 3 (see §73 for a proof). Thus nn decreases as n increases from 3 onwards, and, as it is always greater than unity, it tends to a limit which is greater than or equal to unity. But if nn l, where l > 1, then n > ln, which is certainly untrue for sufficiently large values of n, since ln/n + with n (Exs. 7, 8).]

12. n!n +. [However large Δ may be, n! > Δn if n is large enough. For if un = Δn/n! then un+1/un = Δ/(n + 1), which tends to zero as n , so that un does the same (Ex. 6).]

13. Show that if 1 < x < 1 then

un = m(m 1)(m n + 1) n! xn = m n xn
tends to zero as n .

[If m is a positive integer, un = 0 for n > m. Otherwise

un+1 un = m n n + 1 x x,
unless x = 0.]

73. The limit of 1 + 1 nn. A more difficult problem which can be solved by the help of § 69 arises when φ(n) = {1 + 1/n}n.

It follows from the binomial theorem43 that

(1 + 1 n)n = 1 + n 1 n + n(n 1) 1 2 1 n2 + + n(n 1)(n n + 1) 1 2n 1 nn = 1 + 1 + 1 1 2(1 1 n) + 1 1 2 3(1 1 n)(1 2 n) + + 1 1 2n(1 1 n)(1 2 n)(1 n 1 n ).

The (p + 1)th term in this expression, viz.

1 1 2p 1 1 n 1 2 n 1 p 1 n ,
is positive and an increasing function of n, and the number of terms also increases with n. Hence 1 + 1 nn increases with n, and so tends to a limit or to  + , as n .

But

1 + 1 nn < 1 + 1 + 1 1 2 + 1 1 2 3 + + 1 1 2 3n < 1 + 1 + 1 2 + 1 22 + + 1 2n1 < 3.

Thus 1 + 1 nn cannot tend to  + , and so

lim n1 + 1 nn = e,
where e is a number such that 2 < e 3.

74. Some algebraical lemmas. It will be convenient to prove at this stage a number of elementary inequalities which will be useful to us later on.

(i) It is evident that if α > 1 and r is a positive integer then

rαr > αr1 + αr2 + + 1.
Multiplying both sides of this inequality by α 1, we obtain
rαr(α 1) > αr 1;
and adding r(αr 1) to each side, and dividing by r(r + 1), we obtain

αr+1 1 r + 1 > αr 1 r (α > 1). (1)

Similarly we can prove that

1 βr+1 r + 1 < 1 βr r (0 < β < 1). (2)

It follows that if r and s are positive integers, and r > s, then

αr 1 r > as 1 s ,1 βr r < 1 βs s . (3)

Here 0 < β < 1 < α. In particular, when s = 1, we have

αr 1 > r(α 1),1 βr < r(1 β). (4)

(ii) The inequalities (3) and (4) have been proved on the supposition that r and s are positive integers. But it is easy to see that they hold under the more general hypothesis that r and s are any positive rational numbers. Let us consider, for example, the first of the inequalities (3). Let r = a/b, s = c/d, where ab, cd are positive integers; so that ad > bc. If we put α = γbd, the inequality takes the form

(γad 1)/ad > (γbc 1)/bc;
and this we have proved already. The same argument applies to the remaining inequalities; and it can evidently be proved in a similar manner that

αs 1 < s(α 1),1 βs > s(1 β), (5)

if s is a positive rational number less than 1.

(iii) In what follows it is to be understood that all the letters denote positive numbers, that r and s are rational, and that α and r are greater than 1β and s less than 1. Writing 1/β for α, and 1/α for β, in (4), we obtain

αr 1 < rαr1(α 1),1 βr > rβr1(1 β). (6)

Similarly, from (5), we deduce

αs 1 > sαs1(α 1),1 βs < sβs1(1 β). (7)

Combining (4) and (6), we see that

rαr1(α 1) > αr 1 > r(α 1). (8)

Writing x/y for α, we obtain

rxr1(x y) > xr yr > ryr1(x y) (9)

if x > y > 0. And the same argument, applied to (5) and (7), leads to

sxs1(x y) < xs ys < sys1(x y). (10)

Examples XXVIII. 1. Verify (9) for r = 23, and (10) for s = 1 21 3.

2. Show that (9) and (10) are also true if y > x > 0.

3. Show that (9) also holds for r < 0. [See Chrystal’s Algebra, vol. ii, pp. 43–45.]

4. If φ(n) l, where l > 0, as n , then φk lk, k being any rational number.

[We may suppose that k > 0, in virtue of Theorem III of §66; and that 1 2l < φ < 2l, as is certainly the case from a certain value of n onwards. If k > 1,

kφk1(φ l) > φk lk > klk1(φ l)
or
klk1(l φ) > lk φk > kφk1(l φ),
according as φ > l or φ < l. It follows that the ratio of φk lk and φ llies between k(1 2l)k1 and k(2l)k1. The proof is similar when 0 < k < 1. The result is still true when l = 0, if k > 0.]

5. Extend the results of Exs. XXVII. 7, 8, 9 to the case in which r or k are any rational numbers.

75. The limit of n(xn 1). If in the first inequality (3) of §74 we put r = 1/(n 1), s = 1/n, we see that

(n 1)(αn 1 1) > n(αn 1)
when α > 1. Thus if φ(n) = n(αn 1) then φ(n) decreases steadily as n increases. Also φ(n) is always positive. Hence φ(n) tends to a limit l as n , and l 0.

Again if, in the first inequality (7) of §74, we put s = 1/n, we obtain

n(αn 1) > αn 1 1 α > 1 1 α.
Thus l 1 (1/α) > 0. Hence, if α > 1, we have
limnn(αn 1) = f(α),
where f(α) > 0.

Next suppose β < 1, and let β = 1/α; then n(βn 1) = n(α 1)/αn. Now n(αn 1) f(α), and (Exs. XXVII. 10)

αn 1.
Hence, if β = 1/α < 1, we have
n(βn 1) f(α).
Finally, if x = 1, then n(xn 1) = 0 for all values of n.

Thus we arrive at the result: the limit

limn(xn 1)
defines a function of x for all positive values of x. This function f(x) possesses the properties
f(1/x) = f(x),f(1) = 0,
and is positive or negative according as x > 1 or x < 1.
Later on we shall be able to identify this function with the Napierian logarithm of x.

Example. Prove that f(xy) = f(x) + f(y). [Use the equations

f(xy) = limn(xyn 1) = lim{n(xn 1)yn + n(yn 1)}.]

76. Infinite Series. Suppose that u(n) is any function of n defined for all values of n. If we add up the values of u(ν) for ν = 1, 2, … n, we obtain another function of n, viz.

s(n) = u(1) + u(2) + + u(n),
also defined for all values of n. It is generally most convenient to alter our notation slightly and write this equation in the form
sn = u1 + u2 + + un,
or, more shortly,
sn = ν=1nu ν.

If now we suppose that sn tends to a limit s when n tends to , we have

lim n ν=1nu ν = s.
This equation is usually written in one of the forms
ν=1u ν = s,u1 + u2 + u3 + = s,
the dots denoting the indefinite continuance of the series of u’s.

The meaning of the above equations, expressed roughly, is that by adding more and more of the u’s together we get nearer and nearer to the limit s. More precisely, if any small positive number ε is chosen, we can choose n0(ε) so that the sum of the first n0(ε) terms, or any of greater number of terms, lies between s ε and s + ε; or in symbols

s ε < sn < s + ε,
if n n0(ε). In these circumstances we shall call the series
u1 + u2 +
a convergent infinite series, and we shall call s the sum of the series, or the sum of all the terms of the series.

Thus to say that the series u1 + u2 + converges and has the sum s, or converges to the sum s or simply converges to s, is merely another way of stating that the sum sn = u1 + u2 + + un of the first n terms tends to the limit s as n , and the consideration of such infinite series introduces no new ideas beyond those with which the early part of this chapter should already have made the reader familiar. In fact the sum sn is merely a function φ(n), such as we have been considering, expressed in a particular form. Any function φ(n) may be expressed in this form, by writing

φ(n) = φ(1) + {φ(2) φ(1)} + + {φ(n) φ(n 1)};
and it is sometimes convenient to say that φ(n) converges (instead of ‘tends’) to the limit l, say, as n .

If sn + or sn , we shall say that the series u1 + u2 + is divergent or diverges to  + , or  , as the case may be. These phrases too may be applied to any function φ(n): thus if φ(n) + we may say that φ(n) diverges to  + . If sn does not tend to a limit or to  + or to  , then it oscillates finitely or infinitely: in this case we say that the series u1 + u2 + oscillates finitely or infinitely.44

77. General theorems concerning infinite series. When we are dealing with infinite series we shall constantly have occasion to use the following general theorems.

(1) If u1 + u2 + is convergent, and has the sum s, then a + u1 + u2 + is convergent and has the sum a + s. Similarly a + b + c + + k + u1 + u2 + is convergent and has the sum a + b + c + + k + s.

(2) If u1 + u2 + is convergent and has the sum s, then um+1 + um+2 + is convergent and has the sum

s u1 u2 um.

(3) If any series considered in (1) or (2) diverges or oscillates, then so do the others.

(4) If u1 + u2 + is convergent and has the sum s, then ku1 + ku2 + is convergent and has the sum ks.

(5) If the first series considered in (4) diverges or oscillates, then so does the second, unless k = 0.

(6) If u1 + u2 + and v1 + v2 + are both convergent, then the series (u1 + v1) + (u2 + v2) + is convergent and its sum is the sum of the first two series.

All these theorems are almost obvious and may be proved at once from the definitions or by applying the results of §§ 6366 to the sum sn = u1 + u2 + + un. Those which follow are of a somewhat different character.

(7) If u1 + u2 + is convergent, then lim un = 0.

For un = sn sn1, and sn and sn1 have the same limit s. Hence lim un = s s = 0.

The reader may be tempted to think that the converse of the theorem is true and that if limun = 0 then the series  un must be convergent. That this is not the case is easily seen from an example. Let the series be

1 + 1 2 + 1 3 + 1 4 +
so that un = 1/n. The sum of the first four terms is
1 + 1 2 + 1 3 + 1 4 > 1 + 1 2 + 2 4 = 1 + 1 2 + 1 2.
The sum of the next four terms is 1 5 + 1 6 + 1 7 + 1 8 > 4 8 = 1 2; the sum of the next eight terms is greater than 8 16 = 1 2, and so on. The sum of the first
4 + 4 + 8 + 16 + + 2n = 2n+1
terms is greater than
2 + 1 2 + 1 2 + 1 2 + + 1 2 = 1 2(n + 3),
and this increases beyond all limit with n: hence the series diverges to  + .

(8) If u1 + u2 + u3 + is convergent, then so is any series formed by grouping the terms in brackets in any way to form new single terms, and the sums of the two series are the same.

The reader will be able to supply the proof of this theorem. Here again the converse is not true. Thus 1 1 + 1 1 + oscillates, while

(1 1) + (1 1) +
or 0 + 0 + 0 + converges to 0.

(9) If every term un is positive orzero, then the series  un must either converge or diverge to  + . If it converges, its sum must be positive (unless all the terms are zero, when of course its sum is zero).

For sn is an increasing function of n, according to the definition of § 69, and we can apply the results of that section to sn.

(10)If every term un is positive orzero, then the necessary and sufficient condition that the series  un should be convergent is that it should be possible to find a number K such that the sum of any number of terms is less than K; and, if K can be so found, then the sum of the series is not greater than K.

This also follows at once from § 69. It is perhaps hardly necessary to point out that the theorem is not true if the condition that every un is positive is not fulfilled. For example

1 1 + 1 1 +
obviously oscillates, sn being alternately equal to 1 and to 0.

(11)If u1 + u2 + , v1 + v2 + are two series of positive orzero terms, and the second series is convergent, and if un Kvn, where K is a constant, for all values of n, then the first series is also convergent, and its sum is less than or equal to K times that of the second.

For if v1 + v2 + = t then v1 + v2 + + vn t for all values of n, and so u1 + u2 + + un Kt; which proves the theorem.

Conversely, if un is divergent, and vn Kun, then vn is divergent.

78. The infinite geometrical series. We shall now consider the ‘geometrical’ series, whose general term is un = rn1. In this case

sn = 1 + r + r2 + + rn1 = (1 rn)/(1 r),
except in the special case in which r = 1, when
sn = 1 + 1 + + 1 = n.
In the last case sn +. In the general case sn will tend to a limit if and only if rn does so. Referring to the results of § 72 we see that the series 1 + r + r2 + is convergent and has the sum 1/(1 r) if and only if 1 < r < 1.

If r 1, then sn n, and so sn +; i.e. the series diverges to  + . If r = 1, then sn = 1 or sn = 0 according as n is odd or even: i.e. sn oscillates finitely. If r < 1, then sn oscillates infinitely. Thus, to sum up, the series 1 + r + r2 + diverges to  + if r 1, converges to 1/(1 r) if 1 < r < 1, oscillates finitely if r = 1, and oscillates infinitely if r < 1.

Examples XXIX. 1. Recurring decimals. The commonest example of an infinite geometric series is given by an ordinary recurring decimal. Consider, for example, the decimal .21713¯. This stands, according to the ordinary rules of arithmetic, for

2 10 + 1 102 + 7 103 + 1 104 + 3 105 + 1 106 + 3 107 + = 217 1000 + 13 105 1 1 102 = 2687 12,375.
The reader should consider where and how any of the general theorems of §77 have been used in this reduction.

2. Show that in general

.a1a2amα1α2αn¯ = a1a2amα1αn a1a2an 999000 ,
the denominator containing n 9’s and m 0’s.

3. Show that a pure recurring decimal is always equal to a proper fraction whose denominator does not contain 2 or 5 as a factor.

4. A decimal with m non-recurring and n recurring decimal figures is equal to a proper fraction whose denominator is divisible by 2m or 5m but by no higher power of either.

5. The converses of Exs. 3, 4 are also true. Let r = p/q, and suppose first that q is prime to 10. If we divide all powers of 10 by q we can obtain at most q different remainders. It is therefore possible to find two numbers n1 and n2, where n1 > n2, such that 10n1 and 10n2 give the same remainder. Hence 10n1 10n2 = 10n2(10n1n2 1) is divisible by q, and so 10n 1, where n = n1 n2, is divisible by q. Hence r may be expressed in the form P/(10n 1), or in the form

P 10n + P 102n + ,
i.e. as a pure recurring decimal with n figures. If on the other hand q = 2α5βQ, where Q is prime to 10, and m is the greater of α and β, then 10mr has a denominator prime to 10, and is therefore expressible as the sum of an integer and a pure recurring decimal. But this is not true of 10μr, for any value of μ less than m; hence the decimal for r has exactly m non-recurring figures.

6. To the results of Exs. 2–5 we must add that of Ex. I. 3. Finally, if we observe that

.9¯ = 9 10 + 9 102 + 9 103 + = 1,
we see that every terminating decimal can also be expressed as a mixed recurring decimal whose recurring part is composed entirely of 9’s. For example, .217 = .2169¯. Thus every proper fraction can be expressed as a recurring decimal, and conversely.

7. Decimals in general. The expression of irrational numbers as non-recurring decimals. Any decimal, whether recurring or not, corresponds to a definite number between 0 and 1. For the decimal .a1a2a3a4 stands for the series

a1 10 + a2 102 + a3 103 + .
Since all the digits ar are positive, the sum sn of the first n terms of this series increases with n, and it is certainly not greater than .9¯ or 1. Hence sn tends to a limit between 0 and 1.

Moreover no two decimals can correspond to the same number (except in the special case noticed in Ex. 6). For suppose that .a1a2a3, .b1b2b3 are two decimals which agree as far as the figures ar1br1, while ar > br. Then ar br + 1 > br.br+1br+2 (unless br+1, br+2, … are all 9’s), and so

.a1a2arar+1 > .b1b2brbr+1.
It follows that the expression of a rational fraction as a recurring decimal (Exs. 2–6) is unique. It also follows that every decimal which does not recur represents some irrational number between 0 and 1. Conversely, any such number can be expressed as such a decimal. For it must lie in one of the intervals
0,1/10;1/10,2/10;;9/10,1.
If it lies between r/10 and (r + 1)/10, then the first figure is r. By subdividing this interval into 10 parts we can determine the second figure; and so on. But (Exs. 3, 4) the decimal cannot recur. Thus, for example, the decimal 1.414, obtained by the ordinary process for the extraction of 2, cannot recur.

8. The decimals .1010010001000010 and .2020020002000020, in which the number of zeros between two 1’s or 2’s increases by one at each stage, represent irrational numbers.

9. The decimal .11101010001010, in which the nth figure is 1 if n is prime, and zero otherwise, represents an irrational number. [Since the number of primes is infinite the decimal does not terminate. Nor can it recur: for if it did we could determine m and p so that mm + p, m + 2p, m + 3p, … are all prime numbers; and this is absurd, since the series includes m + mp.]45

Examples XXX. 1. The series rm + rm+1 + is convergent if 1 < r < 1, and its sum is 1/(1 r) 1 r rm1 (§77, (2)).

2. The series rm + rm+1 + is convergent if 1 < r < 1, and its sum is rm/(1 r) (§77, (4)). Verify that the results of Exs. 1 and 2 are in agreement.

3. Prove that the series 1 + 2r + 2r2 + is convergent, and that its sum is (1 + r)/(1 r), (α) by writing it in the form 1 + 2(1 + r + r2 + ), (β) by writing it in the form 1 + 2(r + r2 + ), (γ) by adding the two series 1 + r + r2 + , r + r2 + . In each case mention which of the theorems of §77 are used in your proof.

4. Prove that the ‘arithmetic’ series

a + (a + b) + (a + 2b) +
is always divergent, unless both a and b are zero. Show that, if b is not zero, the series diverges to  + or to  according to the sign of b, while if b = 0 it diverges to  + or  according to the sign of a.

5. What is the sum of the series

(1 r) + (r r2) + (r2 r3) +
when the series is convergent? [The series converges only if 1 < r 1. Its sum is 1, except when r = 1, when its sum is 0.]

6. Sum the series

r2 + r2 1 + r2 + r2 (1 + r2)2 + .
[The series is always convergent. Its sum is 1 + r2, except when r = 0, when its sum is 0.]

7. If we assume that 1 + r + r2 + is convergent then we can prove that its sum is 1/(1 r) by means of §77, (1) and (4). For if 1 + r + r2 + = s then

s = 1 + r(1 + r2 + ) = 1 + rs.

8. Sum the series

r + r 1 + r + r (1 + r)2 +
when it is convergent. [The series is convergent if 1 < 1/(1 + r) < 1, i.e. if r < 2 or if r > 0, and its sum is 1 + r. It is also convergent when r = 0, when its sum is 0.]

9. Answer the same question for the series

r r 1+r + r (1+r)2 ,r + r 1r + r (1r)2 + , 1 r 1+r + r 1+r 2 , 1 + r 1r + r 1r 2 + .

10. Consider the convergence of the series

(1 + r) + (r2 + r3) + , (1 + r + r2) + (r3 + r4 + r5) + , 1 2r + r2 + r3 2r4 + r5 + ,(1 2r + r2) + (r3 2r4 + r5) + ,

and find their sums when they are convergent.

11. If 0 an 1 then the series a0 + a1r + a2r2 + is convergent for 0 r < 1, and its sum is not greater than 1/(1 r).

12. If in addition the series a0 + a1 + a2 + is convergent, then the series a0 + a1r + a2r2 + is convergent for 0 r 1, and its sum is not greater than the lesser of a0 + a1 + a2 + and 1/(1 r).

13. The series

1 + 1 1 + 1 1 2 + 1 1 2 3 +
is convergent. [For 1/(1 2n) 1/2n1.]

14. The series

1 + 1 1 2 + 1 1 2 3 4 + ,1 1 + 1 1 2 3 + 1 1 2 3 4 5 +
are convergent.

15. The general harmonic series

1 a + 1 a + b + 1 a + 2b + ,
where a and b are positive, diverges to  + .

[For un = 1/(a + nb) > 1/{n(a + b)}. Now compare with 1 + 1 2 + 1 3 + .]

16. Show that the series

(u0 u1) + (u1 u2) + (u2 u3) +
is convergent if and only if un tends to a limit as n .

17. If u1 + u2 + u3 + is divergent then so is any series formed by grouping the terms in brackets in any way to form new single terms.

18. Any series, formed by taking a selection of the terms of a convergent series of positive terms, is itself convergent.

79. The representation of functions of a continuous real variable by means of limits. In the preceding sections we have frequently been concerned with limits such as

lim nφn(x),
and series such as
u1(x) + u2(x) + = lim n{u1(x) + u2(x) + + un(x)},
in which the function of n whose limit we are seeking involves, besides n, another variable x. In such cases the limit is of course a function of x. Thus in § 75 we encountered the function
f(x) = lim nn(xn 1) :
and the sum of the geometrical series 1 + x + x2 + is a function of x, viz. the function which is equal to 1/(1 x) if 1 < x < 1 and is undefined for all other values of x.

Many of the apparently ‘arbitrary’ or ‘unnatural’ functions considered in Ch. II are capable of a simple representation of this kind, as will appear from the following examples.

Examples XXXI. 1. φn(x) = x. Here n does not appear at all in the expression of φn(x), and φ(x) = limφn(x) = x for all values of x.

2. φn(x) = x/n. Here φ(x) = limφn(x) = 0 for all values of x.

3. φn(x) = nx. If x > 0, φn(x) +; if x < 0, φn(x) : only when x = 0 has φn(x) a limit (viz. 0) as n . Thus φ(x) = 0 when x = 0 and is not defined for any other value of x.

4. φn(x) = 1/nx, nx/(nx + 1).

5. φn(x) = xn. Here φ(x) = 0, ( 1 < x < 1); φ(x) = 1, (x = 1); and φ(x) is not defined for any other value of x.

6. φn(x) = xn(1 x). Here φ(x) differs from the φ(x) of Ex. 5 in that it has the value 0 when x = 1.

7. φn(x) = xn/n. Here φ(x) differs from the φ(x) of Ex. 6 in that it has the value 0 when x = 1 as well as when x = 1.

8. φn(x) = xn/(xn + 1). [φ(x) = 0, ( 1 < x < 1); φ(x) = 1 2, (x = 1); φ(x) = 1, (x < 1 or x > 1); and φ(x) is not defined when x = 1.]

9. φn(x) = xn/(xn 1), 1/(xn + 1), 1/(xn 1), 1/(xn + xn), 1/(xn xn).

10. φn(x) = (xn 1)/(xn + 1), (nxn 1)/(nxn + 1), (xn n)/(xn + n). [In the first case φ(x) = 1 when x> 1, φ(x) = 1 when x< 1, φ(x) = 0 when x = 1 and φ(x) is not defined when x = 1. The second and third functions differ from the first in that they are defined both when x = 1 and when x = 1: the second has the value 1 and the third the value  1 for both these values of x.]

11. Construct an example in which φ(x) = 1, (x> 1); φ(x) = 1, (x< 1); and φ(x) = 0, (x = 1 and x = 1).

12. φn(x) = x{(x2n 1)/(x2n + 1)}2, n/(xn + xn + n).

13. φn(x) = {xnf(x) + g(x)}/(xn + 1). [Here φ(x) = f(x), (x> 1); φ(x) = g(x), (x< 1); φ(x) = 1 2{f(x) + g(x)}, (x = 1); and φ(x) is undefined when x = 1.]

14. φn(x) = (2/π)arctan(nx). [φ(x) = 1, (x > 0); φ(x) = 0, (x = 0); φ(x) = 1, (x < 0). This function is important in the Theory of Numbers, and is usually denoted by  sgnx.]

15. φn(x) = sinnxπ. [φ(x) = 0 when x is an integer; and φ(x) is otherwise undefined (Ex. XXIV. 7).]

16. If φn(x) = sin(n!xπ) then φ(x) = 0 for all rational values of x (Ex. XXIV. 14). [The consideration of irrational values presents greater difficulties.]

17. φn(x) = (cos2xπ)n. [φ(x) = 0 except when x is integral, when φ(x) = 1.]

18. If N 1752 then the number of days in the year N A.D. is

lim{365 + (cos21 4Nπ)n (cos2 1 100Nπ)n + (cos2 1 400Nπ)n}.

80. The bounds of a bounded aggregate. Let S be any system or aggregate of real numbers s. If there is a number K such that s K for every s of S, we say that S is bounded above. If there is a number k such that s k for every s, we say that S is bounded below. If S is both bounded above and bounded below, we say simply that S is bounded.

Suppose first that S is bounded above (but not necessarily below). There will be an infinity of numbers which possess the property possessed by K; any number greater than K, for example, possesses it. We shall prove that among these numbers there is a least,46 which we shall call M. This number M is not exceeded by any member of S, but every number less than M is exceeded by at least one member of S.

We divide the real numbers ξ into two classes L and R, putting ξ into L or R according as it is or is not exceeded by members of S. Then every ξ belongs to one and one only of the classes L and R. Each class exists; for any number less than any member of S belongs to L, while K belongs to R. Finally, any member of L is less than some member of S, and therefore less than any member of R. Thus the three conditions of Dedekind’s Theorem (§17) are satisfied, and there is a number M dividing the classes.

The number M is the number whose existence we had to prove. In the first place, M cannot be exceeded by any member of S. For if there were such a member s of S, we could write s = M + η, where η is positive. The number M + 1 2η would then belong to L, because it is less than s, and to R, because it is greater than M; and this is impossible. On the other hand, any number less than M belongs to L, and is therefore exceeded by at least one member of S. Thus M has all the properties required.

This number M we call the upper bound of S, and we may enunciate the following theorem. Any aggregate S which is bounded above has an upper bound M. No member of S exceeds M; but any number less than M is exceeded by at least one member of S.

In exactly the same way we can prove the corresponding theorem for an aggregate bounded below (but not necessarily above). Any aggregate S which is bounded below has a lower bound m. No member of S is less than m; but there is at least one member of S which is less than any number greater than m.

It will be observed that, when S is bounded above, M K, and when S is bounded below, m k. When S is bounded, k m M K.

81. The bounds of a bounded function. Suppose that φ(n) is a function of the positive integral variable n. The aggregate of all the values φ(n) defines a set S, to which we may apply all the arguments of §80. If S is bounded above, or bounded below, or bounded, we say that φ(n) is bounded above, or bounded below, or bounded. If φ(n) is bounded above, that is to say if there is a number K such that φ(n) K for all values of n, then there is a number M such that

(i) φ(n) M for all values of n;

(ii) if ε is any positive number then φ(n) > M ε for at least one value of n. This number M we call the upper bound of φ(n). Similarly, if φ(n) is bounded below, that is to say if there is a number k such that φ(n) k for all values of n, then there is a number m such that

(i) φ(n) m for all values of n;

(ii) if ε is any positive number then φ(n) < m + ε for at least one value of n. This number m we call the lower bound of φ(n).

If K exists, M K; if k exists, m k; and if both k and K exist then

k m M K.

82. The limits of indetermination of a bounded function. Suppose that φ(n) is a bounded function, and M and m its upper and lower bounds. Let us take any real number ξ, and consider now the relations of inequality which may hold between ξ and the values assumed by φ(n) for large values of n. There are three mutually exclusive possibilities:

(1) ξ φ(n) for all sufficiently large values of n;

(2) ξ φ(n) for all sufficiently large values of n;

(3) ξ < φ(n) for an infinity of values of n, and also ξ > φ(n) for an infinity of values of n.

In case (1) we shall say that ξ is a superior number, in case (2) that it is an inferior number, and in case (3) that it is an intermediate number. It is plain that no superior number can be less than m, and no inferior number greater than M.

Let us consider the aggregate of all superior numbers. It is bounded below, since none of its members are less than m, and has therefore a lower bound, which we shall denote by Λ. Similarly the aggregate of inferior numbers has an upper bound, which we denote by λ.

We call Λ and λ respectively the upper and lower limits of indetermination of φ(n) as n tends to infinity; and write

Λ = limsupφ(n),λ = liminf φ(n).
These numbers have the following properties:

(1) m λ Λ M;

(2) Λ and λ are the upper and lower bounds of the aggregate of intermediate numbers, if any such exist;

(3) if ε is any positive number, then φ(n) < Λ + ε for all sufficiently large values of n, and φ(n) > Λ ε for an infinity of values of n;

(4) similarly φ(n) > λ ε for all sufficiently large values of n, and φ(n) < λ + ε for an infinity of values of n;

(5) the necessary and sufficient condition that φ(n) should tend to a limit is that Λ = λ, and in this case the limit is l, the common value of λ and Λ.

Of these properties, (1) is an immediate consequence of the definitions; and we can prove (2) as follows. If Λ = λ = l, there can be at most one intermediate number, viz. l, and there is nothing to prove. Suppose then that Λ > λ. Any intermediate number ξ is less than any superior and greater than any inferior number, so that λ ξ Λ. But if λ < ξ < Λ then ξ must be intermediate, since it is plainly neither superior nor inferior. Hence there are intermediate numbers as near as we please to either λ or Λ.

To prove (3) we observe that Λ + ε is superior and Λ ε intermediate or inferior. The result is then an immediate consequence of the definitions; and the proof of (4) is substantially the same.

Finally (5) may be proved as follows. If Λ = λ = l, then

l ε < φ(n) < l + ε
for every positive value of ε and all sufficiently large values of n, so that φ(n) l. Conversely, if φ(n) l, then the inequalities above written hold for all sufficiently large values of n. Hence l ε is inferior and l + ε superior, so that
λ l ε,Λ l + ε,
and therefore Λ λ 2ε. As Λ λ 0, this can only be true if Λ = λ.

Examples XXXII. 1. Neither Λ nor λ is affected by any alteration in any finite number of values of φ(n).

2. If φ(n) = a for all values of n, then m = λ = Λ = M = a.

3. If φ(n) = 1/n, then m = λ = Λ = 0 and M = 1.

4. If φ(n) = (1)n, then m = λ = 1 and Λ = M = 1.

5. If φ(n) = (1)n/n, then m = 1, λ = Λ = 0, M = 1 2.

6. If φ(n) = (1)n{1 + (1/n)}, then m = 2, λ = 1, Λ = 1, M = 3 2.

7. Let φ(n) = sinnθπ, where θ > 0. If θ is an integer then m = λ = Λ = M = 0. If θ is rational but not integral a variety of cases arise. Suppose, e.g., that θ = p/q, p and q being positive, odd, and prime to one another, and q > 1. Then φ(n) assumes the cyclical sequence of values

sin(pπ/q),sin(2pπ/q),,sin{(2q 1)pπ/q},sin(2qpπ/q),.
It is easily verified that the numerically greatest and least values of φ(n) are cos(π/2q) and cos(π/2q), so that
m = λ = cos(π/2q),Λ = M = cos(π/2q).
The reader may discuss similarly the cases which arise when p and q are not both odd.

The case in which θ is irrational is more difficult: it may be shown that in this case m = λ = 1 and Λ = M = 1. It may also be shown that the values of φ(n) are scattered all over the interval [1,1] in such a way that, if ξ is any number of the interval, then there is a sequence n1, n2, … such that φ(nk) ξ as k .47

The results are very similar when φ(n) is the fractional part of nθ.

83. The general principle of convergence for a bounded function. The results of the preceding sections enable us to formulate a very important necessary and sufficient condition that a bounded function φ(n) should tend to a limit, a condition usually referred to as the general principle of convergence to a limit.

THEOREM 1. The necessary and sufficient condition that a bounded function φ(n) should tend to a limit is that, when any positive number ε is given, it should be possible to find a number n0(ε) such that

φ(n2) φ(n1)< ε
for all values of n1 and n2 such that n2 > n1 n0(ε).

In the first place, the condition is necessary. For if φ(n) l then we can find n0 so that

l 1 2ε < φ(n) < l + 1 2ε
when n n0, and so

φ(n2) φ(n1)< ε (1)

when n1 n0 and n2 n0.

In the second place, the condition is sufficient. In order to prove this we have only to show that it involves λ = Λ. But if λ < Λ then there are, however small ε may be, infinitely many values of n such that φ(n) < λ + ε and infinitely many such that φ(n) > Λ ε; and therefore we can find values of n1 and n2, each greater than any assigned number n0, and such that

φ(n2) φ(n1) > Λ λ 2ε,
which is greater than 1 2(Λ λ) if ε is small enough. This plainly contradicts the inequality (1). Hence λ = Λ, and so φ(n) tends to a limit.

84. Unbounded functions. So far we have restricted ourselves to bounded functions; but the ‘general principle of convergence’ is the same for unbounded as for bounded functions, and the words ‘a bounded function’ may be omitted from the enunciation of Theorem 1.

In the first place, if φ(n) tends to a limit l then it is certainly bounded; for all but a finite number of its values are less than l + ε and greater than l ε.

In the second place, if the condition of Theorem 1 is satisfied, we have

φ(n2) φ(n1)< ε
whenever n1 n0 and n2 n0. Let us choose some particular value n1 greater than n0. Then
φ(n1) ε < φ(n2) < φ(n1) + ε
when n2 n0. Hence φ(n) is bounded; and so the second part of the proof of the last section applies also.

The theoretical importance of the ‘general principle of convergence’ can hardly be overestimated. Like the theorems of §69, it gives us a means of deciding whether a function φ(n) tends to a limit or not, without requiring us to be able to tell beforehand what the limit, if it exists, must be; and it has not the limitations inevitable in theorems of such a special character as those of §69. But in elementary work it is generally possible to dispense with it, and to obtain all we want from these special theorems. And it will be found that, in spite of the importance of the principle, practically no applications are made of it in the chapters which follow.48 We will only remark that, if we suppose that

φ(n) = sn = u1 + u2 + + un,
we obtain at once a necessary and sufficient condition for the convergence of an infinite series, viz:

THEOREM 2. The necessary and sufficient condition for the convergence of the series u1 + u2 + is that, given any positive number ε, it should be possible to find n0 so that

un1+1 + un1+2 + + un2< ε
for all values of n1 and n2 such that n2 > n1 n0.

85. Limits of complex functions and series of complex terms. In this chapter we have, up to the present, concerned ourselves only with real functions of n and series all of whose terms are real. There is however no difficulty in extending our ideas and definitions to the case in which the functions or the terms of the series are complex.

Suppose that φ(n) is complex and equal to

ρ(n) + iσ(n),
where ρ(n)σ(n) are real functions of n. Then if ρ(n) and σ(n) converge respectively to limits r and s as n , we shall say that φ(n) converges to the limit l = r + is, and write
lim φ(n) = l.
Similarly, when un is complex and equal to vn + iwn, we shall say that the series
u1 + u2 + u3 +
is convergent and has the sum l = r + is, if the series
v1 + v2 + v3 + ,w1 + w2 + w3 +
are convergent and have the sums rs respectively
.

To say that u1 + u2 + u3 + is convergent and has the sum l is of course the same as to say that the sum

sn = u1 + u2 + + un = (v1 + v2 + + vn) + i(w1 + w2 + + wn)
converges to the limit l as n .

In the case of real functions and series we also gave definitions of divergence and oscillation, finite or infinite. But in the case of complex functions and series, where we have to consider the behaviour both of ρ(n) and of σ(n), there are so many possibilities that this is hardly worth while. When it is necessary to make further distinctions of this kind, we shall make them by stating the way in which the real or imaginary parts behave when taken separately.

86. The reader will find no difficulty in proving such theorems as the following, which are obvious extensions of theorems already proved for real functions and series.

(1) If lim φ(n) = l then lim φ(n + p) = l for any fixed value of p.

(2) If u1 + u2 + is convergent and has the sum l, then a + b + c + + k + u1 + u2 + is convergent and has the sum a + b + c + + k + l, and up+1 + up+2 + is convergent and has the sum l u1 u2 up.

(3) If lim φ(n) = l and lim ψ(n) = m, then

lim{φ(n) + ψ(n)} = l + m.

(4) If lim φ(n) = l, then lim kφ(n) = kl.

(5) If lim φ(n) = l and lim ψ(n) = m, then lim φ(n)ψ(n) = lm.

(6) If u1 + u2 + converges to the sum l, and v1 + v2 + to the sum m, then (u1 + v1) + (u2 + v2) + converges to the sum l + m.

(7) If u1 + u2 + converges to the sum l then ku1 + ku2 + converges to the sum kl.

(8) If u1 + u2 + u3 + is convergent then lim un = 0.

(9) If u1 + u2 + u3 + is convergent, then so is any series formed by grouping the terms in brackets, and the sums of the two series are the same.

As an example, let us prove theorem (5). Let

φ(n) = ρ(n) + iσ(n),ψ(n) = ρ(n) + iσ(n),l = r + is,m = r+ is.

Then

ρ(n) r,σ(n) s,ρ(n) r,σ(n) s.

But

φ(n)ψ(n) = ρρσσ+ i(ρσ+ ρσ),
and
ρρσσrrss,ρσ+ ρσ rs+ rs;
so that
φ(n)ψ(n) rrss+ i(rs+ rs),
i.e.
φ(n)ψ(n) (r + is)(r+ is) = lm.

The following theorems are of a somewhat different character.

(10)In order that φ(n) = ρ(n) + iσ(n) should converge to zero as n , it is necessary and sufficient that

φ(n) = {ρ(n)}2 + {σ(n)}2
should converge to zero.

If ρ(n) and σ(n) both converge to zero then it is plain that ρ2 + σ2 does so. The converse follows from the fact that the numerical value of ρ or σ cannot be greater than ρ2 + σ2.

(11) More generally, in order that φ(n) should converge to a limit l, it is necessary and sufficient that

φ(n) l
should converge to zero.

For φ(n) l converges to zero, and we can apply (10).

(12)Theorems 1 and 2 of §§ 8384 are still true when φ(n) and un are complex.

We have to show that the necessary and sufficient condition that φ(n) should tend to l is that

φ(n2) φ(n1)< ε (1)

when n2 > n1 n0.

If φ(n) l then ρ(n) r and σ(n) s, and so we can find numbers n and ndepending on ε and such that

ρ(n2) ρ(n1)< 1 2ε,σ(n2) σ(n1)< 1 2ε,
the first inequality holding when n2 > n1 n, and the second when n2 > n1 n. Hence
φ(n2) φ(n1)ρ(n2) ρ(n1)+ σ(n2) σ(n1)< ε
when n2 > n1 n0, where n0 is the greater of n and n. Thus the condition (1) is necessary. To prove that it is sufficient we have only to observe that
ρ(n2) ρ(n1)φ(n2) φ(n1)< ε
when n2 > n1 n0. Thus ρ(n) tends to a limit r, and in the same way it may be shown that σ(n) tends to a limit s.

87. The limit of zn as n , z being any complex number. Let us consider the important case in which φ(n) = zn. This problem has already been discussed for real values of z in § 72.

If zn l then zn+1 l, by (1) of § 86. But, by (4) of § 86,

zn+1 = zzn zl,
and therefore l = zl, which is only possible if (al = 0 or (bz = 1. If z = 1 then lim zn = 1. Apart from this special case the limit, if it exists, can only be zero.

Now if z = r(cos θ + i sin θ), where r is positive, then

zn = rn(cos nθ + i sin nθ),
so that zn = rn. Thus zn tends to zero if and only if r < 1; and it follows from (10) of § 86 that
lim zn = 0
if and only if r < 1. In no other case does zn converge to a limit, except when z = 1 and zn 1.

88. The geometric series 1 + z + z2 + when z is complex. Since

sn = 1 + z + z2 + + zn1 = (1 zn)/(1 z),
unless z = 1, when the value of sn is n, it follows that the series 1 + z + z2 + is convergent if and only if r = z < 1. And its sum when convergent is 1/(1 z).

Thus if z = r(cos θ + i sin θ) = r Cis θ, and r < 1, we have

1 + z + z2 + = 1/(1 r Cis θ),  or 1 + r Cis θ + r2 Cis 2θ + = 1/(1 r Cis θ) = (1 r cos θ + ir sin θ)/(1 2r cos θ + r2).

Separating the real and imaginary parts, we obtain

1 + r cos θ + r2 cos 2θ + = (1 r cos θ)/(1 2r cos θ + r2), r sin θ + r2 sin 2θ + = r sin θ/(1 2r cos θ + r2),

provided r < 1. If we change θ into θ + π, we see that these results hold also for negative values of r numerically less than 1. Thus they hold when 1 < r < 1.

Examples XXXIII. 1. Prove directly that φ(n) = rn cosnθ converges to 0 when r < 1 and to 1 when r = 1 and θ is a multiple of 2π. Prove further that if r = 1 and θ is not a multiple of 2π, then φ(n) oscillates finitely; if r > 1 and θ is a multiple of 2π, then φ(n) +; and if r > 1 and θ is not a multiple of 2π, then φ(n) oscillates infinitely.

2. Establish a similar series of results for φ(n) = rn sinnθ.

3. Prove that

zm + zm+1 + = zm/(1 z), zm + 2zm+1 + 2zm+2 + = zm(1 + z)/(1 z),

if and only if z< 1. Which of the theorems of §86 do you use?

4. Prove that if 1 < r < 1 then

1 + 2rcosθ + 2r2 cos2θ + = (1 r2)/(1 2rcosθ + r2).

5. The series

1 + z 1 + z + z 1 + z2 +
converges to the sum 1 1 z 1 + z = 1 + z if z/(1 + z)< 1. Show that this condition is equivalent to the condition that z has a real part greater than  1 2.

Miscellaneous Examples on Chapter IV.

1. The function φ(n) takes the values 1, 0, 0, 0, 1, 0, 0, 0, 1, … when n = 0, 1, 2, …. Express φ(n) in terms of n by a formula which does not involve trigonometrical functions. [φ(n) = 1 4{1 + (1)n + in + (i)n}.]

2. If φ(n) steadily increases, and ψ(n) steadily decreases, as n tends to , and if ψ(n) > φ(n) for all values of n, then both φ(n) and ψ(n) tend to limits, and limφ(n) limψ(n). [This is an immediate corollary from §69.]

3. Prove that, if

φ(n) = 1 + 1 nn,ψ(n) = 1 1 nn,
then φ(n + 1) > φ(n) and ψ(n + 1) < ψ(n). [The first result has already been proved in §73.]

4. Prove also that ψ(n) > φ(n) for all values of n: and deduce (by means of the preceding examples) that both φ(n) and ψ(n) tend to limits as n tends to .49

5. The arithmetic mean of the products of all distinct pairs of positive integers whose sum is n is denoted by Sn. Show that lim(Sn/n2) = 1/6.

(Math. Trip. 1903.)

6. Prove that if x1 = 1 2{x + (A/x)}, x2 = 1 2{x1 + (A/x1)}, and so on, x and A being positive, then limxn = A.

[Prove first that xn A xn + A =(x A x + A)2n .]

7. If φ(n) is a positive integer for all values of n, and tends to  with n, then xφ(n) tends to 0 if 0 < x < 1 and to  + if x > 1. Discuss the behaviour of xφ(n), as n , for other values of x.

8.50If an increases or decreases steadily as n increases, then the same is true of (a1 + a2 + + an)/n.

9. If xn+1 = k + xn, and k and x1 are positive, then the sequence x1x2, x3, … is an increasing or decreasing sequence according as x1 is less than or greater than α, the positive root of the equation x2 = x + k; and in either case xn α as n .

10. If xn+1 = k/(1 + xn), and k and x1 are positive, then the sequences x1x3, x5, … and x2x4, x6, … are one an increasing and the other a decreasing sequence, and each sequence tends to the limit α, the positive root of the equation x2 + x = k.

11. The function f(x) is increasing and continuous (see Ch. V) for all values of x, and a sequence x1x2, x3, … is defined by the equation xn+1 = f(xn). Discuss on general graphical grounds the question as to whether xn tends to a root of the equation x = f(x). Consider in particular the case in which this equation has only one root, distinguishing the cases in which the curve y = f(x) crosses the line y = x from above to below and from below to above.

12. If x1x2 are positive and xn+1 = 1 2(xn + xn1), then the sequences x1x3, x5, … and x2x4, x6, … are one a decreasing and the other an increasing sequence, and they have the common limit 1 3(x1 + 2x2).

13. Draw a graph of the function y defined by the equation

y = limnx2n sin 1 2πx + x2 x2n + 1 .
(Math. Trip. 1901.)

14. The function

y = limn 1 1 + nsin2πx
is equal to 0 except when x is an integer, and then equal to 1. The function
y = limnψ(x) + nφ(x)sin2πx 1 + nsin2πx
is equal to φ(x) unless x is an integer, and then equal to ψ(x).

15. Show that the graph of the function

y = limnxnφ(x) + xnψ(x) xn + xn
is composed of parts of the graphs of φ(x) and ψ(x), together with (as a rule) two isolated points. Is y defined when (ax = 1, (bx = 1, (cx = 0?

16. Prove that the function y which is equal to 0 when x is rational, and to 1 when x is irrational, may be represented in the form

y = limm sgn{sin2(m!πx)}, where
 sgnx = limn(2/π)arctan(nx), as in Ex. XXXI. 14. [If x is rational then sin2(m!πx), and therefore  sgn{sin2(m!πx)}, is equal to zero from a certain value of m onwards: if x is irrational then sin2(m!πx) is always positive, and so  sgn{sin2(m!πx)} is always equal to 1.]

Prove that y may also be represented in the form

1 limm[limn{cos(m!πx)}2n].

17. Sum the series

1 1 ν(ν + 1), 1 1 ν(ν + 1)(ν + k).

[Since

1 ν(ν + 1)(ν + k) = 1 k{ 1 ν(ν + 1)(ν + k 1) 1 (ν + 1)(ν + 2)(ν + k)},
we have
1n 1 ν(ν + 1)(ν + k) = 1 k 1 1 2k 1 (n + 1)(n + 2)(n + k)
and so
1 1 ν(ν + 1)(ν + k) = 1 k(k!).]

18. If z< α, then

L z α = L α 1 + z α + z2 α2 + ;  and if z> α, then L z α = L z 1 + α z + α2 z2 + .

19. Expansion of (Az + B)/(az2 + 2bz + c) in powers of z. Let αβ be the roots of az2 + 2bz + c = 0, so that az2 + 2bz + c = a(z α)(z β). We shall suppose that AB, abc are all real, and α and β unequal. It is then easy to verify that

Az + B az2 + 2bz + c = 1 a(α β) Aα + B z α Aβ + B z β .
There are two cases, according as b2 > ac or b2 < ac.

(1) If b2 > ac then the roots αβ are real and distinct. If z is less than either α or βwe can expand 1/(z α) and 1/(z β) in ascending powers of z (Ex. 18). If z is greater than either α or βwe must expand in descending powers of z; while if z lies between α and β one fraction must be expanded in ascending and one in descending powers of z. The reader should write down the actual results. If z is equal to α or β then no such expansion is possible.

(2) If b2 < ac then the roots are conjugate complex numbers (Ch. III §43), and we can write

α = ρCisφ,β = ρCis(φ),
where ρ2 = αβ = c/a, ρcosφ = 1 2(α + β) = b/a, so that cosφ = b2/ac, sinφ = 1 (b2/ac).

If z< ρ then each fraction may be expanded in ascending powers of z. The coefficient of zn will be found to be

Aρsinnφ + Bsin{(n + 1)φ} aρn+1 sinφ .
If z> ρ we obtain a similar expansion in descending powers, while if z= ρ no such expansion is possible.

20. Show that if z< 1 then

1 + 2z + 3z2 + + (n + 1)zn + = 1/(1 z)2.

[The sum to n terms is 1 zn (1 z)2 nzn 1 z.]

21. Expand L/(z α)2 in powers of z, ascending or descending according as z< α or z> α.

22. Show that if b2 = ac and az< bthen

Az + B az2 + 2bz + c = 0p nzn,
where pn = {(a)n/bn+2}{(n + 1)aB nbA}; and find the corresponding expansion, in descending powers of z, which holds when az> b.

23. Verify the result of Ex. 19 in the case of the fraction 1/(1 + z2). [We have 1/(1 + z2) = zn sin{1 2(n + 1)π}= 1 z2 + z4 .]

24. Prove that if z< 1 then

1 1 + z + z2 = 2 3 0zn sin{2 3(n + 1)π}.

25. Expand (1 + z)/(1 + z2), (1 + z2)/(1 + z3) and (1 + z + z2)/(1 + z4) in ascending powers of z. For what values of z do your results hold?

26. If a/(a + bz + cz2) = 1 + p1z + p2z2 + then

1 + p12z + p 22z2 + = a + cz a cz a2 a2 (b2 2ac)z + c2z2.
(Math. Trip. 1900.)

27. If limnsn = l then

limns1 + s2 + + sn n = l.

[Let sn = l + tn. Then we have to prove that (t1 + t2 + + tn)/n tends to zero if tn does so.

We divide the numbers t1, t2, …, tn into two sets t1, t2, …, tp and tp+1, tp+2, …, tn. Here we suppose that p is a function of n which tends to  as n , but more slowly than n, so that p and p/n 0: e.g. we might suppose p to be the integral part of n.

Let ε be any positive number. However small ε may be, we can choose n0 so that tp+1, tp+2, …, tn are all numerically less than 1 2ε when n n0, and so

(tp+1 + tp+2 + + tn)/n< 1 2ε(n p)/n < 1 2ε.
But, if A is the greatest of the moduli of all the numbers t1, t2, …, we have
(t1 + t2 + + tp)/n< pA/n,
and this also will be less than 1 2ε when n n0, if n0 is large enough, since p/n 0 as n . Thus
(t1 + t2 + + tn)/n(t1 + t2 + + tp)/n+ (tp+1 + + tn)/n< ε
when n n0; which proves the theorem.

The reader, if he desires to become expert in dealing with questions about limits, should study the argument above with great care. It is very often necessary, in proving the limit of some given expression to be zero, to split it into two parts which have to be proved to have the limit zero in slightly different ways. When this is the case the proof is never very easy.

The point of the proof is this: we have to prove that (t1 + t2 + + tn)/n is small when n is large, the t’s being small when their suffixes are large. We split up the terms in the bracket into two groups. The terms in the first group are not all small, but their number is small compared with n. The number in the second group is not small compared with n, but the terms are all small, and their number at any rate less than n, so that their sum is small compared with n. Hence each of the parts into which (t1 + t2 + + tn)/n has been divided is small when n is large.]

28. If φ(n) φ(n 1) l as n , then φ(n)/n l.

[If φ(n) = s1 + s2 + + sn then φ(n) φ(n 1) = sn, and the theorem reduces to that proved in the last example.]

29. If sn = 1 2{1 (1)n}, so that sn is equal to 1 or 0 according as n is odd or even, then (s1 + s2 + + sn)/n 1 2 as n .

[This example proves that the converse of Ex. 27 is not true: for sn oscillates as n .]

30. If cnsn denote the sums of the first n terms of the series

1 2 + cosθ + cos2θ + ,sinθ + sin2θ + ,
then
lim(c1 + c2 + + cn)/n = 0,lim(s1 + s2 + + sn)/n = 1 2 cot 1 2θ.