# 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.

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 $2p > 1$.

• Why not include p = 1/2? – zhw. May 13 '16 at 18:03
• By mistake -- I forgot; I'll add it (currently adding an alternative approach as well). Thanks! – Clement C. May 13 '16 at 18:03
• @ClementC.Great approach. I do have one question though. For $p=\frac{1}{2}$, I was forced to expand $\ln\left(1+x\right)$ up to third order order, since I couldn't use the limit comparison test (implied by Taylor's theorem) with the remainder of the 2nd order expansion. Can you comment about this? – Joshhh May 13 '16 at 18:22
• @Joshhh What was the reason you couldn't use the limit comparison? (I don't see a reason -- may be missing something there). You get $$-\ln\left(1-\frac{(-1)^n}{\sqrt{n}}\right) = \underbrace{\frac{(-1)^n}{\sqrt{n}}}_{a_n} + \underbrace{\frac{1}{2n} + o\left(\frac{1}{n}\right)}_{b_n}.$$ Now, $\sum_n a_n$ converges by Lipschitz/alternating series criterion, and $\sum_n b_n$ diverges by comparison with the Harmonic series (as $b_n \sim_{n\to\infty} \frac{1}{2n}$). – Clement C. May 13 '16 at 18:26
• @ClementC. From Taylor's theorem, the remainder satisfies $\lim\limits_{x\to0}\frac{R(x)}{x^2} = 0$, letting $x = \frac{(-1)^{n+1}}{\sqrt{n}}$, we have that $$\lim\limits_{n\to\infty}\frac{R\left(\frac{(-1)^{n+1}}{\sqrt{n}}\right)}{\frac{1}{n}} = 0$$ and therefore the limit comparison test is inconclusive – Joshhh May 13 '16 at 18:28