Showing divergence of $\sum\limits_{k=1}^{\infty} \log\left(1+\frac{(-1)^{k+1}}{k^\alpha}\right)$ where $0<\alpha<\frac{1}{2}$

Question

I am trying to prove that $$\sum\limits_{k=1}^{\infty} \log\left(1+\frac{(-1)^{k+1}}{k^\alpha}\right)$$ diverges, for any $0<\alpha<\frac{1}{2}$.

I tried showing this by taking the Taylor expansionof $\log(1+\epsilon)$ around $0$ up to order $N$, where $N$ is the minimal integer such that $\alpha N >1$, then substituting $\epsilon = \frac{(-1)^{k+1}}{k^\alpha}$.

This resulted in the sum of the following libnitz serieses + some convergent series (it converges because $\alpha N >1$) + the remainder, and they all converge:

\begin{align*} &\sum\limits_{k=1}^{N} (\log(1+\frac{(-1)^{k+1}}{k^\alpha})\\ =&\sum\frac{(-1)^{k+1}}{k^\alpha}-\sum\frac{1}{2k^{2\alpha}}+\dots+\sum(-1)^{N+1}\frac{(-1)^{k+1}}{k^{\alpha N}\cdot N}+\sum R_N\left(\frac{(-1)^{k+1}}{k^\alpha}\right) \end{align*}

which is clearly in contradiction with the sum of the logs diverging

Answer 1

Use General Leibniz's Test: if $\lim\limits_{n\to\infty}a_n=0$, then $$\sum_{n=1}^\infty a_n\ \mbox{converges}\iff \sum_{n=1}^\infty(a_{2n-1}+a_{2n})\ \mbox{converges},$$ and they have the same sum.

Proof: Let $$S_n=\sum_{k=1}^{n}a_k,\quad T_n=\sum_{k=1}^{n}(a_{2k-1}+a_{2k}),$$ then $$S_{2n}=T_n=S_{2n+1}-a_{2n+1}.$$ this implies they have the same convergence and have the same sum.

The series $$\sum_{n=1}^{\infty}\ln\left(1+\frac{(-1)^{n+1}}{n^\alpha}\right)$$ is convergent if and only if $\alpha>\frac12$.

Proof: By the General Leibniz's Test, the series $\displaystyle\sum_{n=1}^{\infty}\ln\left(1+\frac{(-1)^{n+1}}{n^\alpha}\right)$ and $$\sum_{n=1}^{\infty}\left[\ln\left(1+\frac{1}{(2n-1)^\alpha}\right) +\ln\left(1+\frac{-1}{(2n)^\alpha}\right)\right]$$ have the same convergence. Note that $$\begin{align*} 0\leq\ln\left(1+\frac{1}{(2n-1)^\alpha}\right) +\ln\left(1+\frac{-1}{(2n)^\alpha}\right) &=\ln\frac{((2n-1)^\alpha+1)((2n)^\alpha-1)}{(2n-1)^\alpha(2n)^\alpha}\\ &\sim\frac{(2n)^\alpha-(2n-1)^\alpha-1}{(2n-1)^\alpha(2n)^\alpha}\\ &=\frac{(2n)^\alpha-(2n-1)^\alpha}{(2n-1)^\alpha(2n)^\alpha}-\frac{1}{(2n-1)^\alpha(2n)^\alpha}, \end{align*} $$ and $$\frac{(2n)^\alpha-(2n-1)^\alpha}{(2n-1)^\alpha(2n)^\alpha}\sim\frac{\alpha}{(2n)^{\alpha+1}},\ \frac{1}{(2n-1)^\alpha(2n)^\alpha}\sim\frac{1}{(2n)^{2\alpha}},\ n\to\infty.$$

$\displaystyle\sum_{n=1}^{\infty}\frac{\alpha}{(2n)^{\alpha+1}}$ converges if and only if $\alpha>0$, $\displaystyle\sum_{n=1}^{\infty}\frac{1}{(2n)^{2\alpha}}$ converges if and only if $\alpha>\frac12.$

So $\displaystyle\sum_{n=1}^{\infty}\ln\left(1+\frac{(-1)^{n+1}}{n^\alpha}\right)$ is convergent if and only if $\alpha>\frac12$.

Answer 2

Using $$ \log(1+x)=x+O(x^2) $$ one has $$\sum\limits_{k=1}^{\infty} \log\left(1+\frac{(-1)^{k+1}}{k^\alpha}\right)=\sum\limits_{k=1}^{\infty}\left[\frac{(-1)^{k+1}}{k^\alpha}+O\left(\frac{1}{k^{2\alpha}}\right)\right]. $$ Noting that $\sum\limits_{k=1}^{\infty}\frac{(-1)^{k+1}}{k^\alpha}$ converges for $\alpha>0$ and that $\sum k^{-2\alpha}$ converges if $\alpha>\frac12$ and diverges if $\alpha\le\frac12$, one concludes that the series converges if $\alpha>\frac12$ and diverges if $\alpha\le\frac12$.