186–187. Conditionally convergent series

186. Conditionally convergent series.

B. We have now to consider the second case indicated above, viz. that in which the series of moduli $\sum {\alpha }_n$ diverges to $\mathrm{\infty }$.

In the first place we note that, if $\sum u_n$ is conditionally convergent, then the series $\sum v_n$, $\sum w_n$ of § 184 must both diverge to $\mathrm{\infty }$. For they obviously cannot both converge, as this would involve the convergence of $\sum (v_n+w_n)$ or $\sum {\alpha }_n$. And if one of them, say $\sum w_n$, is convergent, and $\sum v_n$ divergent, then $$\begin{array}{}\text{(1)}& \sum _0^N{u}_n=\sum _0^N{v}_n–\sum _0^N{w}_n,\end{array}$$ and therefore tends to $\mathrm{\infty }$ with $N$, which is contrary to the hypothesis that $\sum u_n$ is convergent.

Hence $\sum v_n$, $\sum w_n$ are both divergent. It is clear from equation (1) above that the sum of a conditionally convergent series is the limit of the difference of two functions each of which tends to $\mathrm{\infty }$ with $n$. It is obvious too that $\sum u_n$ no longer possesses the property of convergent series of positive terms (Ex. XXX. 18), and all absolutely convergent series (Ex. LXXVII. 5), that any selection from the terms itself forms a convergent series. And it seems more than likely that the property prescribed by Dirichlet’s Theorem will not be possessed by conditionally convergent series; at any rate the proof of § 185 fails completely, as it depended essentially on the convergence of $\sum v_n$ and $\sum w_n$ separately. We shall see in a moment that this conjecture is well founded, and that the theorem is not true for series such as we are now considering.

187. Tests of convergence for conditionally convergent series.

It is not to be expected that we should be able to find tests for conditional convergence as simple and general as those of § 167 et seq. It is naturally a much more difficult matter to formulate tests of convergence for series whose convergence, as is shown by equation (1) above, depends essentially on the cancelling of the positive by the negative terms. In the first instance there are no comparison tests for convergence of conditionally convergent series.

For suppose we wish to infer the convergence of $\sum v_n$ from that of $\sum u_n$. We have to compare $$v_0+v_1+\cdots +v_n,u_0+u_1+\cdots +u_n.$$ If every $u$ and every $v$ were positive, and every $v$ less than the corresponding $u$, we could at once infer that $$v_0+v_1+\cdots +v_n<u_0+\cdots +u_n,$$ and so that $\sum v_n$ is convergent. If the $u$’s only were positive and every $v$ numerically less than the corresponding $u$, we could infer that $$|v_0|+|v_1|+\cdots +|v_n|<u_0+\cdots +u_n,$$ and so that $\sum v_n$ is absolutely convergent. But in the general case, when the $u$’s and $v$’s are both unrestricted as to sign, all that we can infer is that $$|v_0|+|v_1|+\cdots +|v_n|<|u_0|+\cdots +|u_n|.$$ This would enable us to infer the absolute convergence of $\sum v_n$ from the absolute convergence of $\sum u_n$; but if $\sum u_n$ is only conditionally convergent we can draw no inference at all.

Example. We shall see shortly that the series $1–\frac{1}{2}+\frac{1}{3}–\frac{1}{4}+\dots$ is convergent. But the series $\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\dots$ is divergent, although each of its terms is numerically less than the corresponding term of the former series.

It is therefore only natural that such tests as we can obtain should be of a much more special character than those given in the early part of this chapter.