## 171. Further tests for convergence and divergence.

The examples on § 168 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 5 these tests failed, and we had to fall back upon a special device, which was in essence that of using the series of 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 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 $\tfrac{2}{3},\quad \tfrac{4}{9},\quad \tfrac{8}{27},\ \dots;\qquad 1,\quad \tfrac{1}{4096},\quad \tfrac{1}{531,\ 441},\ \dots$ whose general terms are $$(\frac{2}{3})^{n}$$ and $$n^{-12}$$, the second seems at first sight to decrease far more rapidly. But this is far from being the case; if only we go far enough into the sequences we shall find the terms of the first sequence very much the smaller. For example, $(2/3)^{4} = 16/81 < 1/5,\quad (2/3)^{12} < (1/5)^{3} < (1/10)^{2},\quad (2/3)^{1000} < (1/10)^{166},$ while $1000^{-12} = 10^{-36};$ so that the $$1000$$th term of the first sequence is less than the $$10^{130}$$th part of the corresponding term of the second sequence. Thus the series $$\sum (2/3)^{n}$$ is far more rapidly convergent than the series $$\sum n^{-12}$$, and even this series is very much more rapidly convergent than $$\sum n^{-2}$$.1

## 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 $$u_{n+1} \leq u_{n}$$ 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.2 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 $$\lim nu_{n} = 0$$.

Suppose that $$nu_{n}$$ does not tend to zero. Then it is possible to find a positive number $$\delta$$ such that $$nu_{n} \geq \delta$$ 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 $$n_{1}u_{n_{1}} \geq \delta$$, $$n_{2}u_{n_{2}} \geq \delta$$, …. But, since $$u_{n}$$ decreases as $$n$$ increases, we have $\begin{gathered} u_{0} + u_{1} + \dots + u_{n_{1} – 1} \geq n_{1}u_{n_{1}} \geq \delta,\\ u_{n_{1}} + \dots + u_{n_{2} – 1} \geq (n_{2} – n_{1})u_{n_{2}} > \tfrac{1}{2} n_{2}u_{n_{2}} \geq \tfrac{1}{2} \delta,\\ u_{n_{2}} + \dots + u_{n_{3} – 1} \geq (n_{3} – n_{2})u_{n_{3}} > \tfrac{1}{2} n_{3}u_{n_{3}} \geq \tfrac{1}{2} \delta,\end{gathered}$ 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 $\delta + \tfrac{1}{2} \delta + \tfrac{1}{2} \delta + \dots;$ and therefore $$\sum u_{n}$$ is divergent.

Example LXIX
1. Use Abel’s theorem to show that $$\sum (1/n)$$ and $$\sum \{1/(an + b)\}$$ are divergent. [Here $$nu_{n} \to 1$$ or $$nu_{n} \to 1/a$$.]

2. Show that Abel’s theorem is not true if we omit the condition that $$u_{n}$$ decreases as $$n$$ increases. [The series $1 + \frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{4} + \frac{1}{5^{2}} + \frac{1}{6^{2}} + \frac{1}{7^{2}} + \frac{1}{8^{2}} + \frac{1}{9} + \frac{1}{10^{2}} + \dots,$ in which $$u_{n} = 1/n$$ or $$1/n^{2}$$, according as $$n$$ is or is not a perfect square, is convergent, since it may be rearranged in the form $\frac{1}{2^{2}} + \frac{1}{3^{2}} + \frac{1}{5^{2}} + \frac{1}{6^{2}} + \frac{1}{7^{2}} + \frac{1}{8^{2}} + \frac{1}{10^{2}} + \dots + \left(1 + \frac{1}{4} + \frac{1}{9} + \dots\right),$ and each of these series is convergent. But, since $$nu_{n} = 1$$ whenever $${n}$$ is a perfect square, it is clearly not true that $$nu_{n} \to 0$$.]

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

[Take the series $$\sum(1/n)$$ 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 $1 + \tfrac{1}{2} \cdot \tfrac{1}{2} + \tfrac{1}{3} \left(\tfrac{1}{3} + \tfrac{1}{4}\right) + \tfrac{1}{4} \left(\tfrac{1}{5} + \tfrac{1}{6} + \tfrac{1}{7} + \tfrac{1}{8}\right) + \dots;$ and this series is divergent, since its terms are greater than those of $1 + \tfrac{1}{2} \cdot \tfrac{1}{2} + \tfrac{1}{3} \cdot \tfrac{1}{2} + \tfrac{1}{4} \cdot \tfrac{1}{2} + \dots,$ which is divergent. But it is easy to see that the terms of the series $1 + \tfrac{1}{2} \cdot \tfrac{1}{2} + \tfrac{1}{3} \cdot\tfrac{1}{3} + \tfrac{1}{3} \cdot \tfrac{1}{4} + \tfrac{1}{4} \cdot \tfrac{1}{5} + \tfrac{1}{4} \cdot \tfrac{1}{6} + \dots$ satisfy the condition that $$nu_{n} \to 0$$. In fact $$nu_{n} = 1/\nu$$ if $$2^{\nu-2} < n \leq 2^{\nu-1}$$, and $$\nu \to \infty$$ as $$n \to \infty$$

## 174. Maclaurin’s (or Cauchy’s) Integral Test.

If $$u_{n}$$ decreases steadily as $$n$$ increases, we can write $$u_{n} = \phi(n)$$ and suppose that $$\phi(n)$$ is the value assumed, when $$x = n$$, by a continuous and steadily decreasing function $$\phi(x)$$ of the continuous variable $$x$$. Then, If $$\nu$$ is any positive integer, we have $\phi(\nu – 1) \geq \phi(x) \geq \phi(\nu)$ when $$\nu – 1 \leq x \leq \nu$$. Let $v_{\nu} = \phi(\nu – 1) – \int_{\nu-1}^{\nu} \phi(x)\, dx = \int_{\nu-1}^{\nu} \{\phi(\nu – 1) – \phi(x)\}\, dx,$ so that $0 \leq v_{\nu} \leq \phi(\nu – 1) – \phi(\nu).$ Then $$\sum v_{\nu}$$ is a series of positive terms, and $v_{2} + v_{3} + \dots + v_{n} \leq \phi(1) – \phi(n) \leq \phi(1).$ Hence $$\sum v_{\nu}$$ is convergent, and so $$v_{2} + v_{3} + \dots + v_{n}$$ or $\sum_{1}^{n-1} \phi(\nu) – \int_{1}^{n} \phi(x)\, dx$ tends to a positive limit as $$n \to \infty$$.

Let us write $\Phi(\xi) = \int_{1}^{\xi} \phi(x)\, dx,$ so that $$\Phi(\xi)$$ is a continuous and steadily increasing function of $$\xi$$. Then $u_{1} + u_{2} + \dots + u_{n-1} – \Phi(n)$ tends to a positive limit, not greater than $$\phi(1)$$, as $$n \to \infty$$. Hence $$\sum u_{\nu}$$ is convergent or divergent according as $$\Phi(n)$$ tends to a limit or to infinity as $$n \to \infty$$, and therefore, since $$\Phi(n)$$ increases steadily, according as $$\Phi(\xi)$$ tends to a limit or to infinity as $$\xi \to \infty$$. Hence

if $$\phi(x)$$ is a function of $$x$$ which is positive and continuous for all values of $$x$$ greater than unity, and decreases steadily as $$x$$ increases, then the series $\phi(1) + \phi(2) + \dots$ does or does not converge according as $\Phi(\xi) = \int_{1}^{\xi} \phi(x)\, dx$ does or does not tend to a limit $$l$$ as $$\xi \to \infty$$; and, in the first case, the sum of the series is not greater than $$\phi(1) + l$$.

The sum must in fact be less than $$\phi(1) + l$$. For it follows from (6) of § 160, and Ch.VII, MiscEx 41, that $$v_{\nu} < \phi(\nu – 1) – \phi(\nu)$$, unless $$\phi(x) = \phi(\nu)$$ throughout the interval $${[\nu – 1, \nu]}$$; and this cannot be true for all values of $$\nu$$.

Example LXX
1. Prove that $\sum_{1}^{\infty} \frac{1}{n^{2} + 1} < \tfrac{1}{2} + \tfrac{1}{4}\pi.$

2. Prove that $-\tfrac{1}{2} \pi < \sum_{1}^{\infty} \frac{a}{a^{2} + n^{2}} < \tfrac{1}{2} \pi.$

3. Prove that if $$m > 0$$ then $\frac{1}{m^{2}} + \frac{1}{(m + 1)^{2}} + \frac{1}{(m + 2)^{2}} + \dots < \frac{m + 1}{m}.$

1. Five terms suffice to give the sum of $$\sum n^{-12}$$ correctly to $$7$$ places of decimals, whereas some $$10,000,000$$ are needed to give an equally good approximation to $$\sum n^{-2}$$. A large number of numerical results of this character will be found in Appendix III (compiled by Mr J. Jackson) to the author’s tract ‘Orders of Infinity’ (Cambridge Math. Tracts, No. 12).↩︎
2. This theorem was discovered by Abel but forgotten, and rediscovered by Pringsheim.↩︎