Chapter VIII
The Convergence of Infinite Series and Infinite Integrals

165. IN Ch. IV we explained what was meant by saying that an infinite series is convergent, divergent, or oscillatory, and illustrated our definitions by a few simple examples, mainly derived from the geometrical series

We shall often use the notation

166. Series of Positive Terms. The theory of the convergence of series is comparatively simple when all the terms of the series considered are positive.⁸⁷ We shall consider such series first, not only because they are the easiest to deal with, but also because the discussion of the convergence of a series containing negative or complex terms can often be made to depend upon a similar discussion of a series of positive terms only.

When we are discussing the convergence or divergence of a series we may disregard any finite number of terms. Thus, when a series contains a finite number only of negative or complex terms, we may omit them and apply the theorems which follow to the remainder.

167. It will be well to recall the following fundamental theorems established in § 77.

A. A series of positive terms must be convergent or diverge to $\infty$, and cannot oscillate.

B. The necessary and sufficient condition that $\sum u_n$ should be convergent is that there should be a number $K$ such that

C. The comparison theorem. If $\sum u_n$ is convergent, and $v_n\leqq u_n$ for all values of $n$, then $\sum v_n$ is convergent, and $\sum v_n\leqq \sum u_n$. More generally, if $v_n\leqq K{u}_n$, where $K$ is a constant, then $\sum v_n$ is convergent and $\sum v_n\leqq K\sum u_n$. And if $\sum u_n$ is divergent, and $v_n\geqq K{u}_n$, then $\sum v_n$ is divergent.⁸⁸

Moreover, in inferring the convergence or divergence of $\sum v_n$ by means of one of these tests, it is sufficient to know that the test is satisfied for sufficiently large values of $n$, i.e. for all values of $n$ greater than a definite value $n_0$. But of course the conclusion that $\sum v_n\leqq K\sum u_n$ does not necessarily hold in this case.

A particularly useful case of this theorem is

D. If $\sum u_n$ is convergent $divergent$ and $u_n/v_n$ tends to a limit other than zero as $n\to \infty$, then $\sum v_n$ is convergent $divergent$.

168. First applications of these tests. The one important fact which we know at present, as regards the convergence of any special class of series, is that $\sum r^n$ is convergent if $r<1$ and divergent if $r\geqq 1$.⁸⁹ It is therefore natural to try to apply Theorem C, taking $u_n=r^n$. We at once find

1. The series $\sum v_n$ is convergent if $v_n\leqq K{r}^n$, where $r<1$, for all sufficiently large values of $n$.

When $K=1$, this condition may be written in the form $v_n^{1/n}\leqq r$. Hence we obtain what is known as Cauchy’s test for the convergence of a series of positive terms; viz.

2. The series $\sum v_n$ is convergent if $v_n^{1/n}\leqq r$, where $r<1$, for all sufficiently large values of $n$.

There is a corresponding test for divergence, viz.

2a. The series $\sum v_n$ is divergent if $v_n^{1/n}\geqq 1$ for an infinity of values of $n$.

This hardly requires proof, for $v_n^{1/n}\geqq 1$ involves $v_n\geqq 1$. The two theorems 2 and 2a are of very wide application, but for some purposes it is more convenient to use a different test of convergence, viz.

3. The series $\sum v_n$ is convergent if $v_{n+1}/v_n\leqq r$, $r<1$, for all sufficiently large values of $n$.

To prove this we observe that if $v_{n+1}/v_n\leqq r$ when $n\geqq n_0$ then

In the simplest cases which occur in analysis it often happens that $v_{n+1}/v_n$ or $v_n^{1/n}$ tends to a limit as $n\to \infty$.⁹⁰ When this limit is less than $1$, it is evident that the conditions of Theorems 2 or 3 above are satisfied. Thus

4. If $v_n^{1/n}$ or $v_{n+1}/v_n$ tends to a limit less than unity as $n\to \infty$, then the series $\sum v_n$ is convergent.

It is almost obvious that if either function tend to a limit greater than unity, then $\sum v_n$ is divergent. We leave the formal proof of this as an exercise to the reader. But when $v_n^{1/n}$ or $v_{n+1}/v_n$ tends to $1$ these tests generally fail completely, and they fail also when $v_n^{1/n}$ or $v_{n+1}/v_n$ oscillates in such a way that, while always less than $1$, it assumes for an infinity of values of $n$ values approaching indefinitely near to $1$. And the tests which involve $v_{n+1}/v_n$ fail even when that ratio oscillates so as to be sometimes less than and sometimes greater than $1$. When $v_n^{1/n}$ behaves in this way Theorem 2a is sufficient to prove the divergence of the series. But it is clear that there is a wide margin of cases in which some more subtle tests will be needed.

Examples LXVII. 1. Apply Cauchy’s and d’Alembert’s tests (as specialised in 4 above) to the series $\sum n^k{r}^n$, where $k$ is a positive rational number.

[Here , so that d’Alembert’s test shows at once that the series is convergent if $r<1$ and divergent if $r>1$. The test fails if $r=1$: but the series is then obviously divergent. Since $lim{n}^{1/n}=1$ (Ex. XXVII. 11), Cauchy’s test leads at once to the same conclusions.]

2. Consider the series . [We may suppose $A$ positive. If the coefficient of $r^n$ is denoted by , then and, by D of §167, the series behaves like $\sum n^k{r}^n$.]

3. Consider

[The series behaves like $\sum n^{k-l}{r}^n$. The case in which $r=1$, $k<l$ requires further consideration.]

4. We have seen (Ch. IV, Misc. Ex. 17) that the series

5. Show that the series $\sum n^{-p}$, where $p$ is an integer not less than $2$, is convergent. [Since , this follows from the convergence of the series considered in Ex. 4. It has already been shown in §77, (7) that the series is divergent if $p=1$, and it is obviously divergent if $p\leqq 0$.]

6. Show that the series

7. If $m_n$ is a positive integer, and $m_{n+1}>m_n$, then the series $\sum r^{m_n}$ is convergent if $r<1$ and divergent if $r\geqq 1$. For example the series $1+r+r^4+r^9+\dots$ is convergent if $r<1$ and divergent if $r\geqq 1$.

8. Sum the series $1+2r+2r^4+\dots$ to $24$ places of decimals when $r=.1$ and to $2$ places when $r=.9$. [If $r=.1$, then the first $5$ terms give the sum $1.2002000020000002$, and the error is

9. If $0<a<b<1$, then the series $a+b+a^2+b^2+a^3+\dots$ is convergent. Show that Cauchy’s test may be applied to this series, but that d’Alembert’s test fails. [For

10. The series $1+r+\frac{r^2}{2!}+\frac{r^3}{3!}+\dots$ and $1+r+\frac{r^2}{2^2}+\frac{r^3}{3^3}+\dots$ are convergent for all positive values of $r$.

11. If $\sum u_n$ is convergent then so are $\sum u_n^2$ and .

12. If $\sum u_n^2$ is convergent then so is $\sum u_n/n$. [For and  is convergent.]

13. Show that

[To prove the first result we note that

by theorems (8) and (6) of §77.]

14. Prove by a reductio ad absurdum that  is divergent. [If the series were convergent we should have, by the argument used in Ex. 13,

169. Before proceeding further in the investigation of tests of convergence and divergence, we shall prove an important general theorem concerning series of positive terms.

Dirichlet’s Theorem.⁹¹ The sum of a series of positive terms is the same in whatever order the terms are taken.

This theorem asserts that if we have a convergent series of positive terms, $u_0+u_1+u_2+\dots$ say, and form any other series

The proof is extremely simple. Let $s$ be the sum of the series of $u\prime$s. Then the sum of any number of terms, selected from the $u\prime$s, is not greater than $s$. But every $v$ is a $u$, and therefore the sum of any number of terms selected from the $v\prime$s is not greater than $s$. Hence $\sum v_n$ is convergent, and its sum $t$ is not greater than $s$. But we can show in exactly the same way that $s\leqq t$. Thus $s=t$.

170. Multiplication of Series of Positive Terms. An immediate corollary from Dirichlet’s Theorem is the following theorem: if $u_0+u_1+u_2+\dots$ and $v_0+v_1+v_2+\dots$ are two convergent series of positive terms, and $s$ and $t$ are their respective sums, then the series

Arrange all the possible products of pairs $u_m{v}_n$ in the form of a doubly infinite array

(1) We begin with the single term $u_0{v}_0$ for which $m+n=0$; then we take the two terms $u_1{v}_0$, $u_0{v}_1$ for which $m+n=1$; then the three terms $u_2{v}_0$, $u_1{v}_1$, $u_0{v}_2$ for which $m+n=2$; and so on. We thus obtain the series

(2) We begin with the single term $u_0{v}_0$ for which both suffixes are zero; then we take the terms $u_1{v}_0$, $u_1{v}_1$, $u_0{v}_1$ which involve a suffix $1$ but no higher suffix; then the terms $u_2{v}_0$, $u_2{v}_1$, $u_2{v}_2$, $u_1{v}_2$, $u_0{v}_2$ which involve a suffix $2$ but no higher suffix; and so on. The sums of these groups of terms are respectively equal to

and the sum of the first $n+1$ groups is

The sum of the series formed in the second manner is $st$. But the first series is (when the brackets are removed) a rearrangement of the second; and therefore, by Dirichlet’s Theorem, it converges to the sum $st$. Thus the theorem is proved.

Examples LXVIII. 1. Verify that if $r<1$ then

2.⁹²If either of the series $u_0+u_1+\dots$, $v_0+v_1+\dots$ is divergent, then so is the series , except in the trivial case in which every term of one series is zero.

3. If the series $u_0+u_1+\dots$, $v_0+v_1+\dots$, $w_0+w_1+\dots$ converge to sums $r$, $s$, $t$, then the series $\sum {\lambda }_k$, where ${\lambda }_k=\sum u_m{v}_n{w}_p$, the summation being extended to all sets of values of $m$, $n$, $p$ such that $m+n+p=k$, converges to the sum $rst$.

4. If $\sum u_n$ and $\sum v_n$ converge to sums $s$ and $t$, then the series $\sum w_n$, where $w_n=\sum u_l{v}_m$, the summation extending to all pairs $l$, $m$ for which $lm=n$, converges to the sum $st$.

171. Further tests for convergence and divergence. The examples on pp. 1049–1054 suffice to show that there are simple and interesting types of series of positive terms which cannot be dealt with by the general tests of § 168. In fact, if we consider the simplest type of series, in which $u_{n+1}/u_n$ tends to a limit as $n\to \infty$, the tests of § 168 generally fail when this limit is $1$. Thus in Ex. LXVII. 5 these tests failed, and we had to fall back upon a special device, which was in essence that of using the series of Ex. LXVII. 4 as our comparison series, instead of the geometric series.

The fact is that the geometric series, by comparison with which the tests of §168 were obtained, is not only convergent but very rapidly convergent, far more rapidly than is necessary in order to ensure convergence. The tests derived from comparison with it are therefore naturally very crude, and much more delicate tests are often wanted.

We proved in Ex. XXVII. 7 that $n^k{r}^n\to 0$ as $n\to \infty$, provided $r<1$, whatever value $k$ may have; and in Ex. LXVII. 1 we proved more than this, viz. that the series $\sum n^k{r}^n$ is convergent. It follows that the sequence $r$, $r^2$, $r^3$, …, $r^n$, …, where $r<1$, diminishes more rapidly than the sequence $1^{-k}$, $2^{-k}$, $3^{-k}$, …, $n^{-k}$, …. This seems at first paradoxical if $r$ is not much less than unity, and $k$ is large. Thus of the two sequences

172. We shall proceed to establish two further tests for the convergence or divergence of series of positive terms, Maclaurin’s (or Cauchy’s) Integral Test and Cauchy’s Condensation Test, which, though very far from being completely general, are sufficiently general for our needs in this chapter.

In applying either of these tests we make a further assumption as to the nature of the function $u_n$, about which we have so far assumed only that it is positive. We assume that $u_n$ decreases steadily with $n$: i.e. that $u_{n+1}\leqq u_n$ for all values of $n$, or at any rate all sufficiently large values.

This condition is satisfied in all the most important cases. From one point of view it may be regarded as no restriction at all, so long as we are dealing with series of positive terms: for in virtue of Dirichlet’s theorem above we may rearrange the terms without affecting the question of convergence or divergence; and there is nothing to prevent us rearranging the terms in descending order of magnitude, and applying our tests to the series of decreasing terms thus obtained.

But before we proceed to the statement of these two tests, we shall state and prove a simple and important theorem, which we shall call Abel’s Theorem.⁹⁴ This is a one-sided theorem in that it gives a sufficient test for divergence only and not for convergence, but it is essentially of a more elementary character than the two theorems mentioned above.

173. Abel’s (or Pringsheim’s) Theorem. If $\sum u_n$ is a convergent series of positive and decreasing terms, then $limn{u}_n=0$.

Suppose that $n{u}_n$ does not tend to zero. Then it is possible to find a positive number $\delta$ such that $n{u}_n\geqq \delta$ for an infinity of values of $n$. Let $n_1$ be the first such value of $n$; $n_2$ the next such value of $n$ which is more than twice as large as $n_1$; $n_3$ the next such value of $n$ which is more than twice as large as $n_2$; and so on. Then we have a sequence of numbers $n_1$, $n_2$, $n_3$, … such that $n_2>2n_1$, $n_3>2n_2$, … and so $n_2-n_1>\frac{1}{2}{n}_2$, $n_3-n_2>\frac{1}{2}{n}_3$, …; and also $n_1{u}_{n_1}\geqq \delta$, $n_2{u}_{n_2}\geqq \delta$, …. But, since $u_n$ decreases as $n$ increases, we have

and so on. Thus we can bracket the terms of the series $\sum u_n$ so as to obtain a new series whose terms are severally greater than those of the divergent series

Examples LXIX. 1. Use Abel’s theorem to show that and are divergent. [Here $n{u}_n\to 1$ or $n{u}_n\to 1/a$.]

2. Show that Abel’s theorem is not true if we omit the condition that $u_n$ decreases as $n$ increases. [The series

3. The converse of Abel’s theorem is not true, i.e. it is not true that, if $u_n$ decreases with $n$ and $limn{u}_n=0$, then $\sum u_n$ is convergent.

[Take the series and multiply the first term by $1$, the second by $\frac{1}{2}$, the next two by $\frac{1}{3}$, the next four by $\frac{1}{4}$, the next eight by $\frac{1}{5}$, and so on. On grouping in brackets the terms of the new series thus formed we obtain

174. Maclaurin’s (or Cauchy’s) Integral Test.⁹⁵ If $u_n$ decreases steadily as $n$ increases, we can write and suppose that  is the value assumed, when $x=n$, by a continuous and steadily decreasing function  of the continuous variable $x$. Then, If $\nu$ is any positive integer, we have

Let us write

The sum must in fact be less than . For it follows from (6) of §160, and Ch. VII, Misc. Ex. 41, that , unless throughout the interval ; and this cannot be true for all values of $\nu$.

Examples LXX. 1. Prove that

2. Prove that

3. Prove that if $m>0$ then

175. The series $\sum n^{-s}$. By far the most important application of the Integral Test is to the series

If $s\leqq 0$ then it is obvious that the series is divergent. If $s>0$ then $u_n$ decreases as $n$ increases, and we can apply the test. Here

So far as divergence for $s<1$ is concerned, this result might have been derived at once from comparison with , which we already know to be divergent.

It is however interesting to see how the Integral Test may be applied to the series , when the preceding analysis fails. In this case

Examples LXXI. 1. Prove by an argument similar to that used above, and without integration, that , where $s<1$, tends to infinity with $\xi$.

2. The series $\sum n^{-2}$, $\sum n^{-3/2}$, $\sum n^{-11/10}$ are convergent, and their sums are not greater than $2$, $3$, $11$ respectively. The series $\sum n^{-1/2}$, $\sum n^{-10/11}$ are divergent.

3. The series , where $a>0$, is convergent or divergent according as $t>1+s$ or $t\leqq 1+s$. [Compare with $\sum n^{s-t}$.]

4. Discuss the convergence or divergence of the series

5. Prove that

6. If then the series is convergent. If then it is divergent.

176. Cauchy’s Condensation Test. The second of the two tests mentioned in § 172 is as follows: if is a decreasing function of $n$, then the series  is convergent or divergent according as  is convergent or divergent.

We can prove this by an argument which we have used already (§ 77) in the special case of the series . In the first place

If diverges then so do and , and then the inequalities just obtained show that  diverges.

On the other hand

For our present purposes the field of application of this test is practically the same as that of the Integral Test. It enables us to discuss the series $\sum n^{-s}$ with equal ease. For $\sum n^{-s}$ will converge or diverge according as $\sum 2^n{2}^{-ns}$ converges or diverges, i.e. according as $s>1$ or $s\leqq 1$.

Examples LXXII. 1. Show that if $a$ is any positive integer greater than $1$ then  is convergent or divergent according as is convergent or divergent. [Use the same arguments as above, taking groups of $a$, $a^2$, $a^3$, … terms.]

2. If converges then it is obvious that . Hence deduce Abel’s Theorem of §173.

177. Infinite Integrals. The Integral Test of § 174 shows that, if  is a positive and decreasing function of $x$, then the series is convergent or divergent according as the integral function  does or does not tend to a limit as $x\to \infty$. Let us suppose that it does tend to a limit, and that

So far we have supposed positive and decreasing. But it is natural to extend our definition to other cases. Nor is there any special point in supposing the lower limit to be unity. We are accordingly led to formulate the following definition:

If  is a function of $t$ continuous when $t\geqq a$, and

is convergent and has the value $l$.

The ordinary integral between limits $a$ and $A$, as defined in Ch. VII, we shall sometimes call in contrast a finite integral.

On the other hand, when

These definitions suggest the following remarks.

(i) If we write

(ii) In the special case in which  is always positive it is clear that  is an increasing function of $x$. Hence the only alternatives are convergence and divergence to $\infty$.

(iii) The integral (1) of course depends on $a$, but is quite independent of $t$, and is in no way altered by the substitution of any other letter for $t$ (cf. §157).

(iv) Of course the reader will not be puzzled by the use of the term infinite integral to denote something which has a definite value such as $2$ or $\frac{1}{2}\pi$. The distinction between an infinite integral and a finite integral is similar to that between an infinite series and a finite series: no one supposes that an infinite series is necessarily divergent.

(v) The integral was defined in §§156 and 157 as a simple limit, i.e. the limit of a certain finite sum. The infinite integral is therefore the limit of a limit, or what is known as a repeated limit. The notion of the infinite integral is in fact essentially more complex than that of the finite integral, of which it is a development.

(vi) The Integral Test of §174 may now be stated in the form: if  is positive and steadily decreases as $x$ increases, then the infinite series and the infinite integral converge or diverge together.

(vii) The reader will find no difficulty in formulating and proving theorems for infinite integrals analogous to those stated in (1)–(6) of §77. Thus the result analogous to (2) is that if is convergent, and $b>a$, then is convergent and

178. The case in which  is positive. It is natural to consider what are the general theorems, concerning the convergence or divergence of the infinite integral (1) of § 177, analogous to theorems A–D of § 167. That A is true of integrals as well as of series we have already seen in § 177, (ii). Corresponding to B we have the theorem that the necessary and sufficient condition for the convergence of the integral (1) is that it should be possible to find a constant $K$ such that

Similarly, corresponding to C, we have the theorem: if is convergent, and for all values of $x$ greater than $a$, then is convergent and

We may observe that d’Alembert’s test (§ 168), depending as it does on the notion of successive terms, has no analogue for integrals; and that the analogue of Cauchy’s test is not of much importance, and in any case could only be formulated when we have investigated in greater detail the theory of the function , as we shall do in Ch. IX. The most important special tests are obtained by comparison with the integral

There is one fundamental property of a convergent infinite series in regard to which the analogy between infinite series and infinite integrals breaks down. If  is convergent then ; but it is not always true, even when  is always positive, that if is convergent then .

Consider for example the function  whose graph is indicated by the thick line in the figure. Here the height of the peaks corresponding to the points $x=1$, $2$, $3$, … is in each case unity, and the breadth of the peak corresponding

to $x=n$ is . The area of the peak is , and it is evident that, for any value of $\xi$,

Examples LXXIII. 1. The integral

2. Which of the integrals ${\int \; <!--nolimits-->}_a^{\infty }\frac{dx}{\sqrt{x}}$, ${\int \; <!--nolimits-->}_a^{\infty }\frac{dx}{x^{4/3}}$,

3. The integrals

4. The integrals

5. Integrals to $-\infty$. If tends to a limit $l$ as $\xi \to -\infty$, then we say that is convergent and equal to $l$. Such integrals possess properties in every respect analogous to those of the integrals discussed in the preceding sections: the reader will find no difficulty in formulating them.

6. Integrals from $-\infty$ to $+\infty$. If the integrals

7. Prove that

8. Prove generally that

9. Prove that if is convergent then .