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 tends to a limit as , the tests of § 168 generally fail when this limit is . 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 as , provided , whatever value may have; and in Ex LXVII. 1 we proved more than this, viz. that the series is convergent. It follows that the sequence , , , …, , …, where , diminishes more rapidly than the sequence , , , …, , …. This seems at first paradoxical if is not much less than unity, and is large. Thus of the two sequences whose general terms are and , 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, while so that the th term of the first sequence is less than the th part of the corresponding term of the second sequence. Thus the series is far more rapidly convergent than the series , and even this series is very much more rapidly convergent than .
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 , about which we have so far assumed only that it is positive. We assume that decreases steadily with : i.e. that for all values of , 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. 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 is a convergent series of positive and decreasing terms, then .
Suppose that does not tend to zero. Then it is possible to find a positive number such that for an infinity of values of . Let be the first such value of ; the next such value of which is more than twice as large as ; the next such value of which is more than twice as large as ; and so on. Then we have a sequence of numbers , , , … such that , , … and so , , …; and also , , …. But, since decreases as increases, we have and so on. Thus we can bracket the terms of the series so as to obtain a new series whose terms are severally greater than those of the divergent series and therefore is divergent.
Example LXIX
1. Use Abel’s theorem to show that
and
are divergent. [Here
or
.]
2. Show that Abel’s theorem is not true if we omit the condition that decreases as increases. [The series in which or , according as is or is not a perfect square, is convergent, since it may be rearranged in the form and each of these series is convergent. But, since whenever is a perfect square, it is clearly not true that .]
3. The converse of Abel’s theorem is not true, i.e. it is not true that, if decreases with and , then is convergent.
[Take the series
and multiply the first term by
, the second by
, the next two by
, the next four by
, the next eight by
, and so on. On grouping in brackets the terms of the new series thus formed we obtain
and this series is divergent, since its terms are greater than those of
which is divergent. But it is easy to see that the terms of the series
satisfy the condition that
. In fact
if
, and
as
174. Maclaurin’s (or Cauchy’s) Integral Test.
If decreases steadily as increases, we can write and suppose that is the value assumed, when , by a continuous and steadily decreasing function of the continuous variable . Then, If is any positive integer, we have when . Let so that Then is a series of positive terms, and Hence is convergent, and so or tends to a positive limit as .
Let us write so that is a continuous and steadily increasing function of . Then tends to a positive limit, not greater than , as . Hence is convergent or divergent according as tends to a limit or to infinity as , and therefore, since increases steadily, according as tends to a limit or to infinity as . Hence
if is a function of which is positive and continuous for all values of greater than unity, and decreases steadily as increases, then the series does or does not converge according as does or does not tend to a limit as ; and, in the first case, the sum of the series is not greater than .
The sum must in fact be less than . For it follows from (6) of § 160, and Ch.VII, MiscEx 41, that , unless throughout the interval ; and this cannot be true for all values of .
Example LXX
1. Prove that
2. Prove that
3. Prove that if then