Is $\sum\limits_{n=2}^{\infty} \ln \left[1+\frac{(-1)^{n}}{n^{p}}\right](p>0)$ convergent? Is $S=\sum\limits_{n=2}^{\infty} \ln \left[1+\dfrac{(-1)^{n}}{n^{p}}\right](p>0)$ convergent?
I can prove the case when $p\geq 1$: If $p>1$, $\sum\limits_{n=2}^{\infty} \left|\ln \left[1+\dfrac{(-1)^{n}}{n^{p}}\right]\right|\leq \sum\limits_{n=2}^{\infty} \left|\dfrac{(-1)^{n}}{n^{p}}\right|=\sum\limits_{n=2}^{\infty}\dfrac{1}{n^{p}}$, which is convergent, so $S$ converges absolutely. If $p=1$, $\sum\limits_{n=2}^{\infty} \ln \left[\dfrac{n+(-1)^{n}}{n}\right]=\sum\limits_{n=2}^{\infty} \ln(n+(-1)^n)-\ln(n)=0 \text{ or } \ln(\dfrac{2n+1}{2n})$ converges. How to deal with $p<1$?
Any suggestions will be greatly appreciated.
 A: Hint:
$$ \ln \left[1+\dfrac{(-1)^{n}}{n^{p}}\right]=\frac{(-1)^n}{n^p}-\frac{1}{2n^{2p}}+o(\frac{1}{n^{2p}}) $$
A: 1) If $\frac{1}{2} < p$, the series $\sum_{n=2}^\infty \ln (1 + \tfrac{(-1)^n}{n^p})$ is convergent.
It is easy to prove that $0 \le x - \ln (1 + x) \le 2x^2$ for $x > -\frac{3}{4}$. 
Note that $\frac{(-1)^n}{n^p} \ge - \frac{1}{3^p} > - \frac{3}{4}$ for $n\ge 2$.
Thus, we have, for $n\ge 2$, 
$$0 \le \frac{(-1)^n}{n^p} - \ln \left(1 + \frac{(-1)^n}{n^p}\right) \le \frac{2}{n^{2p}}.$$
Since $\sum_{n=2}^\infty \frac{2}{n^{2p}}$ is convergent, by comparison test, 
$\sum_{n=2}^\infty \left[\frac{(-1)^n}{n^p} - \ln \left(1 + \frac{(-1)^n}{n^p}\right)\right]$
is convergent. Since $\sum_{n=2}^\infty \frac{(-1)^n}{n^p}$ is convergent (alternating series test),
$\sum_{n=2}^\infty \ln (1 + \tfrac{(-1)^n}{n^p})$ is convergent.
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2) If $0 < p \le \frac{1}{2}$, the series $\sum_{n=1}^\infty \ln (1 + \tfrac{(-1)^n}{n^p})$ is divergent.
It is easy to prove that $\ln (1+x) \le x - \frac{x^2}{4}$ for $- 1 < x < 1$. 
Note that $-1 < \frac{(-1)^n}{n^p} < 1$ for $n \ge 2$.
Thus, we have, for $n\ge 2$,
$$\ln \left(1 + \frac{(-1)^n}{n^p}\right) \le \frac{(-1)^n}{n^p} - \frac{1}{4n^{2p}}.$$
Denote $S_N = \sum_{n=2}^N \ln (1 + \tfrac{(-1)^n}{n^p})$. We have
$$S_N \le \sum_{n=2}^N \frac{(-1)^n}{n^p} - \sum_{n=2}^N \frac{1}{4n^{2p}}.$$
Note that $\sum_{n=2}^\infty \frac{(-1)^n}{n^p}$ is convergent.
Also, $\lim_{N\to \infty} \sum_{n=2}^N \frac{1}{4n^{2p}} = \infty$.
Thus,
$$\lim_{N\to \infty} \left(\sum_{n=2}^N \frac{(-1)^n}{n^p} - \sum_{n=2}^N \frac{1}{4n^{2p}}\right)
= -\infty.$$
Thus, $\lim_{N\to\infty} S_N = -\infty$.
