TITLE: convergence of alternating series — weakening a hypothesis
QUESTION [6 upvotes]: A comment below this answer inspires this question.
Suppose $a_n\in\mathbb{R}$ for $n=1,2,3,\ldots$ and $|a_n|\to0$ as $n\to\infty$.
Further suppose the terms alternate in sign.
If moreover the sequence $\{|a_n|\}_{n=1}^\infty$ is decreasing, then $\displaystyle\sum_{n=1}^\infty a_n$ converges.
How much can the hypothesis that it is decreasing be weakened while still being strong enough that the sum must converge?  And are there any interesting or useful weaker hypotheses?

REPLY [3 votes]: Here is one of my favorite counter-examples for when you don't assume that the sequence $(|a_n|)$ is decreasing. Let us put, for all $n \geq 2$,
$$a_n := \ln \left( 1 + \frac{(-1)^n}{\sqrt{n}} \right).$$
This sequence converges to $0$, and is alternating. However, since \ln (1+x) = $x-x^2/2 + O (x^3)$, we get:
$$a_n := \frac{(-1)^n}{\sqrt{n}} - \frac{1}{2} \left( \frac{(-1)^n}{\sqrt{n}} \right)^2 + O (n^{-\frac{3}{2}}) = \frac{(-1)^n}{\sqrt{n}} - \frac{1}{2n} + O (n^{-\frac{3}{2}}).$$
The series whose general term is $\frac{(-1)^n}{\sqrt{n}}$ is convergent, since it is alternating. The $O (n^{-\frac{3}{2}})$ term is summable, by comparison with Riemann sums. What is left is $\frac{1}{2n}$, whose corresponding series is divergent. Hence, $\sum_{k=0}^{n-1} a_k$ diverges to $- \infty$.
More generally, "alternating but not summable" + "non-negative sequence, which decays faster but is still not summable" gives a sequence which is equivalent to the initial alternating sequence, but whose sum does not converges. Something like $\frac{(-1)^n}{\sqrt{n}} + \frac{1}{n}$ is a typical example (but I prefer the sequence $(a_n)$, where the trap is concealed - it shows that you have to be careful).