## 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 $$\infty$$.

If $$\sum u_{n}$$ is convergent, but $$\sum |u_{n}|$$ divergent, the original series is said to be conditionally convergent.

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 $$\infty$$. 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 $\begin{equation*} \sum_{0}^{N} u_{n} = \sum_{0}^{N} v_{n} – \sum_{0}^{N} w_{n}, \tag{1} \end{equation*}$ and therefore tends to $$\infty$$ 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 $$\infty$$ 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 $v_{0} + v_{1} + \dots + v_{n},\quad u_{0} + u_{1} + \dots + u_{n}.$ If every $$u$$ and every $$v$$ were positive, and every $$v$$ less than the corresponding $$u$$, we could at once infer that $v_{0} + v_{1} + \dots + v_{n} < u_{0} + \dots + u_{n},$ 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 $|v_{0}| + |v_{1}| + \dots + |v_{n}| < u_{0} + \dots + u_{n},$ 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 $|v_{0}| + |v_{1}| + \dots + |v_{n}| < |u_{0}| + \dots + |u_{n}|.$ 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.