Convergence of $\sum\log\left(1-\frac{(-1)^n}{n^p}\right)$ I am trying to find whether the following sum converges or diverges:
$$
\log{2} + \log\left(1-\frac{1}{2^p}\right) + \log\left(1+\frac{1}{3^p}\right) + \ldots = \sum\limits_{n=1}^{\infty}\log\left(1-\frac{(-1)^n}{n^p}\right)
$$
I have discovered that this series converge for $p\geq 1$, diverge for $p\leq 0$, and also diverges for $p = \frac{1}{2}$.
My guess is that the series diverge for $0<p<1$. Here, the ones that answered claimed that it's convergent for $0<p\leq 1$, which is wrong for $p=\frac{1}{2}$, and also one of the answers used the limit comparison test, which is invalid for series with mixed-sign terms.
I have tried to use the Taylor series, but I am not familiar with any theorem stating that I can switch the order of infinite sums. Also, the ratio test is inconclusive.
 A: Method 1:
Using Taylor expansions (not quite series, only the local, asymptotic version to few terms): when $n\to \infty$, for $p > 0$,
$$
 -\ln\left(1-\frac{(-1)^n}{n^p}\right)
= \frac{(-1)^n}{n^p} + \frac{1}{2n^{2p}} + o\left(\frac{1}{n^{2p}}\right)
$$
The first term is the general term of a (conditionally) convergent series by the alternating series criterion. The second, $$
b_n = \frac{1}{2n^{2p}} + o\left(\frac{1}{n^{2p}}\right)
$$
is the general term of a series that will be (absolutely) convergent for $2p > 1$ by the comparison test and divergent otherwise (comparing to the series $\sum_n \frac{1}{n^{2p}}$).
So, to sum up:


*

*if $p>\frac{1}{2}$, the series converges as the sum of a (conditionally) convergent and an (absolutely) convergent series.

*if $0 < p \leq \frac{1}{2}$, the series diverges as the sum of a (conditionally) convergent and a divergent series.



Method 2: Still Taylor expansion, (but in a different fashion).
Write $S_{N} = \sum_{n=1}^N \ln\left(1-\frac{(-1)^n}{n^p}\right)$, for $N\geq 1$.
First, since $S_{2N+1} - S_{2N} = \ln\left(1+\frac{1}{(2N+1)^p}\right) \xrightarrow[N\to\infty]{} 0$, it is sufficient to study the convergence of $(S_{2N})_N$.
Let us group the terms by consecutive pairs:
$$\begin{align}
S_{2N} &= \sum_{n=1}^N \left(
\ln\left(1+\frac{1}{(2n-1)^p}\right)+\ln\left(1-\frac{1}{(2n)^p}\right) \right)
\\
&= \sum_{n=1}^N 
\ln\left(\left( 1+\frac{1}{(2n-1)^p}\right)\left(1-\frac{1}{(2n)^p}\right) \right)\\
&= \sum_{n=1}^N 
\ln\left( 1+\frac{1}{(2n-1)^p}-\frac{1}{(2n)^p}-\frac{1}{(2n\cdot(2n-1))^p} \right)
\end{align}$$
Now, this is unpleasant, but to apply theorems of comparisons to this $S_{2N}$, we can do a Taylor expansion (to order 2 or so) of what is inside the logarithm, then do a first-order expansion of the logarithm itself:
$$
 1+\frac{1}{(2n-1)^p}-\frac{1}{(2n)^p}-\frac{1}{(2n\cdot(2n-1))^p}
= 1-\frac{1}{(2n)^{2p}} + o\left(\frac{1}{n^{2p}}\right)
$$
so
$$
 \ln\left( 1+\frac{1}{(2n-1)^p}-\frac{1}{(2n)^p}-\frac{1}{(2n\cdot(2n-1))^p}\right)
= -\frac{1}{(2n)^{2p}} + o\left(\frac{1}{n^{2p}}\right)
$$
and by theorems of comparison (again with a $p$-series) we get that $(S_{2N})_N$ converges iff $2p > 1$.
