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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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Michael Hardy
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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. – mrf Jul 14 '12 at 17:32
  • I'm thinking that being bounded above by a decreasing sequence might not be strong enough. If so, an example would show that. – Michael Hardy Jul 14 '12 at 17:33
  • What actually provoked the question was that the comment said you need to show it's decreasing. "Need" seemed a bit strong. – Michael Hardy Jul 14 '12 at 17:35
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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.)

     – mrf Jul 14 '12 at 17:35

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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).

D. Thomine
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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 $\alpha\leqslant1$.

Did
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