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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?

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    Here's one that's probably neither interesting, nor useful. If $a_k = (-1)^k b_k\,c_k$ with $b_k, c_k > 0$, $c_k$ is decreasing to $0$ and $\sum_{k=1}^n (-1)^k b_k$ is uniformly bounded in $n$, then the original sum converges (by Dirichlet's test). I have no idea how to find $b_k$ and $c_k$ to make this work in practice... In theory the $b_k$ could "absorb" some of the non-monotonicity.2012-07-14
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    I'm thinking that being bounded above by a decreasing sequence might not be strong enough. If so, an example would show that.2012-07-14
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    What actually provoked the question was that the comment said you _need_ to show it's decreasing. "Need" seemed a bit strong.2012-07-14
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    It's not enough. Take $a_k = 1/k$ if $k$ is even and $a_k = -1/k^2$ if $k$ is odd. Then $|a_k| < 1/k$. (Granted, "need" was perhaps a bad choice of words, but conditional convergence is notoriously tricky. Monotonicity is certainly not *required*. If the positive and negative parts converge separately, i.e. if the series is absolutely convergent, it's easy to write down examples where $|a_k|$ is not monotone.)2012-07-14

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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 $\alpha\leqslant1$.

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    But with $c_n=0$ this would be the alternating series test without the decreasing bit?2012-07-14
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    @joriki Thanks.2012-07-14
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    More generally, convergent + convergent is convergent. So the sum of two alternating series may not be alternating (if the signs are opposite), but is convergent.2012-07-14
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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).