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 \[1 + x + x^{2} + \dots\] and other series closely connected with it. In this chapter we shall pursue the subject in a more systematic manner, and prove a number of theorems which enable us to determine when the simplest series which commonly occur in analysis are convergent.

We shall often use the notation \[u_{m} + u_{m+1} + \dots + u_{n} = \sum_{m}^{n} \phi(\nu),\] and write \(\sum\limits_{0}^{\infty} u_{n}\), or simply \(\sum u_{n}\), for the infinite series \(u_{0} + u_{1} + u_{2} + \dots\).1


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.2 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.



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 \[u_{0} + u_{1} + \dots + u_{n} < K\] for all values of \(n\).

C. The comparison theorem. If \(\sum u_{n}\) is convergent, and \(v_{n} \leq u_{n}\) for all values of \(n\), then \(\sum v_{n}\) is convergent, and \(\sum v_{n} \leq \sum u_{n}\). More generally, if \(v_{n} \leq Ku_{n}\), where \(K\) is a constant, then \(\sum v_{n}\) is convergent and \(\sum v_{n} \leq K \sum u_{n}\). And if \(\sum u_{n}\) is divergent, and \(v_{n} \geq Ku_{n}\), then \(\sum v_{n}\) is divergent.3

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 \(\sum v_{n} \leq K \sum u_{n}\) 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 \(r \geq 1\).4 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 \(v_{n} \leq Kr^{n}\), where \(r < 1\), for all sufficiently large values of \(n\).

When \(K = 1\), this condition may be written in the form \(v_{n}^{1/n} \leq r\). 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 \(v_{n}^{1/n} \leq r\), 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 \(v_{n}^{1/n} \geq 1\) for an infinity of values of \(n\).

This hardly requires proof, for \(v_{n}^{1/n} \geq 1\) involves \(v_{n} \geq 1\). 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 \(v_{n+1}/v_{n} \leq r\), \(r < 1\), for all sufficiently large values of \(n\).

To prove this we observe that if \(v_{n+1}/v_{n} \leq r\) when \(n \geq n_{0}\) then \[v_{n} = \frac{v_{n}}{v_{n-1}}\, \frac{v_{n-1}}{v_{n-2}} \dots \frac{v_{n_{0}+1}}{v_{n_{0}}}\, v_{n_{0}} \leq \frac{v_{n_{0}}}{r^{n_{0}}} r^{n};\] and the result follows by comparison with the convergent series \(\sum r^{n}\). This test is known as d’Alembert’s test. We shall see later that it is less general, theoretically, than Cauchy’s, in that Cauchy’s test can be applied whenever d’Alembert’s can, and sometimes when the latter cannot. Moreover the test for divergence which corresponds to d’Alembert’s test for convergence is much less general than the test given by Theorem 2. It is true, as the reader will easily prove for himself, that if \(v_{n+1}/v_{n} \geq r \geq 1\) for all values of \(n\), or all sufficiently large values, then \(\sum v_{n}\) is divergent. But it is not true (see Ex LXVII. 9) that this is so if only \(v_{n+1}/v_{n} \geq r \geq 1\) for an infinity of values of \(n\), whereas in Theorem 2 our test had only to be satisfied for such an infinity of values. None the less d’Alembert’s test is very useful in practice, because when \(v_{n}\) is a complicated function \(v_{n+1}/v_{n}\) is often much less complicated and so easier to work with.

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 \(n \to \infty\).5 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 2 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.

Example 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 \(v_{n+1}/v_{n} = \{(n + 1)/n\}^{k} r \to r\), 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 \(\sum(An^{k} + Bn^{k-1} + \dots + K) r^{n}\). [We may suppose \(A\) positive. If the coefficient of \(r^{n}\) is denoted by \(P(n)\), then \(P(n)/n^{k} \to A\) and, by D of § 167, the series behaves like \(\sum n^{k} r^{n}\).]

3. Consider \[\sum \frac{An^{k} + Bn^{k-1} + \dots + K} {\alpha n^{l} + \beta n^{l-1} + \dots + \kappa} r^{n}\quad (A > 0,\ \alpha > 0).\]

[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, MiscEx 17) that the series \[\sum \frac{1}{n(n + 1)},\quad \sum \frac{1}{n(n + 1)\dots (n + p)}\] are convergent. Show that Cauchy’s and d’Alembert’s tests both fail when applied to them. [For \(\lim u_{n}^{1/n} = \lim (u_{n+1}/u_{n}) = 1\).]

5. Show that the series \(\sum n^{-p}\), where \(p\) is an integer not less than \(2\), is convergent. [Since \(\lim \{n(n + 1)\dots (n + p – 1)\}/n^{p} = 1\), 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 \(p \leq 0\).]

6. Show that the series \[\sum \frac{An^{k} + Bn^{k-1} + \dots + K} {\alpha n^{l} + \beta n^{l-1} + \dots + \kappa}\] is convergent if \(l > k + 1\) and divergent if \(l \leq k + 1\).

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 \(r \geq 1\). For example the series \(1 + r + r^{4} + r^{9} + \dots\) is convergent if \(r < 1\) and divergent if \(r \geq 1\).

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 \(1.200\ 200\ 002\ 000\ 000\ 2\), and the error is \[2r^{25} + 2r^{36} + \dots < 2r^{25} + 2r^{36} + 2r^{47} + \dots = 2r^{25}/(1 – r^{11}) < 3/10^{25}.\] If \(r = .9\), then the first \(8\) terms give the sum \(5.458\dots\), and the error is less than \(2r^{64}/(1 – r^{17}) < .003\).]

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 \[v_{2n+1}/v_{2n} = (b/a)^{n+1} \to \infty,\quad v_{2n+2}/v_{2n+1} = b(a/b)^{n+2} \to 0.]\]

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

11. If \(\sum u_{n}\) is convergent then so are \(\sum u_{n}^{2}\) and \(\sum u_{n}/(1 + u_{n})\).

12. If \(\sum u_{n}^{2}\) is convergent then so is \(\sum u_{n}/n\). [For \(2u_{n}/n \leq u_{n}^{2} + (1/n^{2})\) and \(\sum (1/n^{2})\) is convergent.]

3. Show that \[1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots = \frac{3}{4}\left(1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots \right)\] and \[1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{6^{2}} + \frac{1}{7^{2}} + \frac{1}{9^{2}} + \dots = \frac{15}{16} \left(1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots\right).\]

[To prove the first result we note that \[\begin{aligned} 1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots &= \left(1 + \frac{1}{2^{2}}\right) + \left(\frac{1}{3^{2}} + \frac{1}{4^{2}}\right) + \dots\\ &= 1 + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \dots + \frac{1}{2^{2}} \left(1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \dots\right),\end{aligned}\] by theorems (8) and (6) of § 77.]

14. Prove by a reductio ad absurdum that \(\sum (1/n)\) is divergent. [If the series were convergent we should have, by the argument used in Ex. 13, \[1 + \tfrac{1}{2} + \tfrac{1}{3} + \dots = (1 + \tfrac{1}{3} + \tfrac{1}{5} + \dots) + \tfrac{1}{2} (1 + \tfrac{1}{2} + \tfrac{1}{3}+ \dots),\] or \[\tfrac{1}{2} + \tfrac{1}{4} + \tfrac{1}{6} + \dots = 1 + \tfrac{1}{3} + \tfrac{1}{5} + \dots\] which is obviously absurd, since every term of the first series is less than the corresponding term of the second.]

  1. It is of course a matter of indifference whether we denote our series by \(u_{1} + u_{2} + \dots\) (as in ) or by \(u_{0} + u_{1} + \dots\) (as in Ch. IV). Later in this chapter we shall be concerned with series of the type \(a_{0} + a_{1}x + a_{2}x^{2} + \dots\): for these the latter notation is clearly more convenient. We shall therefore adopt this as our standard notation. But we shall not adhere to it systematically, and we shall suppose that \(u_{1}\) is the first term whenever this course is more convenient. It is more convenient, for example, when dealing with the series \(1 + \frac{1}{2} + \frac{1}{3} + \dots\), to suppose that \(u_{n} = 1/n\) and that the series begins with \(u_{1}\), than to suppose that \(u_{n} = 1/(n + 1)\) and that the series begins with \(u_{0}\). This remark applies, e.g., to Ex lxviii . 4.↩︎
  2. Here and in what follows ‘positive’ is to be regarded as including zero.↩︎
  3. The last part of this theorem was not actually stated in § 77, but the reader will have no difficulty in supplying the proof.↩︎
  4. We shall use \(r\) in this chapter to denote a number which is always positive or zero.↩︎
  5. It will be proved in Ch. IX (Ex LXXXVII. 36) that if \(v_{n+1}/v_{n} \to l\) then \(v_{n}^{1/n} \to l\). That the converse is not true may be seen by supposing that \(v_{n} = 1\) when \(n\) is odd and \(v_{n} = 2\) when \(n\) is even.↩︎

$\leftarrow$ Chapter VII Main Page 169. Dirichlet’s Theorem $\rightarrow$