convergence of alternating series — weakening a hypothesis 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?
 A: 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).
A: One may ask that $a_n=b_n+c_n$ where $(b_n)$ is alternating (that is, $b_n\to0$ monotonically  and $(-1)^nb_n$ of constant sign) and $(c_n)$ is absolutely summable (that is, $\sum\limits_n|c_n|$ finite). 
This applies readily to show that $\sum\limits_n\dfrac{(-1)^n}{n^\alpha+(-1)^n}$ converges if and only if $\alpha\gt\frac12$. Note that the absolute value of the general term is not monotonous when $\alpha\leqslant1$.
