Convergence of $\sum_{n\geq 2}{\log(1+\frac{(-1)^n}{n^\alpha}})$ Let $\alpha >0$, I want to study the  convergence of the series $\sum_{n\geq 2}{\log(1+\frac{(-1)^n}{n^\alpha}})$. I know that the series  $\sum_{n\geq 2}{\frac{(-1)^n}{n^\alpha}}$ is a convergent series by the alternating series test. Can we say that
$\log(1+\frac{(-1)^n}{n^\alpha})$ is equivalent at $\infty$ to $\frac{(-1)^n}{n^\alpha}$ and then conclude that the series converges?
 A: Note that for $\alpha > 0$ and $n \geq 2$,
$$ \log\left(1 + \frac{(-1)^{n}}{n^{\alpha}} \right) = \frac{(-1)^{n}}{n^{\alpha}} - \frac{1}{2n^{2\alpha}} + O(n^{-3\alpha}). $$
Thus if $\alpha \leq \frac{1}{2}$, then
$$ \sum_{n=2}^{N} \log\left(1 + \frac{(-1)^{n}}{n^{\alpha}} \right) = \sum_{n=2}^{N} \frac{(-1)^{n}}{n^{\alpha}} - \frac{1+o(1)}{2} \sum_{n=2}^{N} \frac{1}{n^{2\alpha}}. \tag{1} $$
diverges to $-\infty$. In fact, $\text{(1)}$ shows that this is an if-and-only-if condition:
$$ \sum_{n=2}^{\infty} \log\left(1 + \frac{(-1)^{n}}{n^{\alpha}} \right) \text{ converges if and only if } \alpha > \frac{1}{2}. $$
A: Your use of "equivalent" is not rigorous, and that turns out to be crucial in this problem.  To use the Alternating Series Test, you need three things:
(i) $\ln(1+\frac{(-1)^n}{n^\alpha}) \to 0$ as $n \to \infty$.
(ii) The terms $\ln(1+\frac{(-1)^n}{n^\alpha})$ alternate in sign.
(iii) $|\ln(1+\frac{(-1)^{n+1}}{(n+1)^\alpha})| < |\ln(1+\frac{(-1)^{n}}{{n}^\alpha})|$ for all $n \geq 1$, or at least for large enough $n$.
(i) and (ii) are obvious. (iii) might take a little work.  Maybe you could consider the cases of even $n$ and odd $n$ separately and us properties of logs (log of product is sum of logs, $\ln$ is increasing, etc.)   I just tried the case where $n$ is even and the inequality in (iii) worked about to be true if $n$ was large enough (which is all you need).
EDIT: I just saw the answer above that concludes the series diverges.  So I looked at (iii) again.  (iii) turns out not to be true if $n$ is odd, for any $\alpha > 0$, even if you assume $n$ is large.  So the Alternating Series Test tells you nothing.
