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

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

• The second sum on the RHS diverges because $2\alpha\lt 1$ Commented Mar 20 at 11:51
• $\log (1+x)=x-x^{2}/2+...+(-1)^{N}\frac {x^{N}}N+o(x^{N+1})$ Choose $N$ such that $\alpha (N+1)>1$. Commented Mar 20 at 11:54
• If the claimed result is true then it is a little bit surprising, because $\log (1+x) \approx x,$ and so $\sum\limits_{k=1}^{\infty} \log\left(1+\frac{(-1)^{k+1}}{k^\alpha}\right) \approx \sum\limits_{k=1}^{\infty} \frac{(-1)^{k+1}}{k^\alpha}$ which converges because it is an alternating sequence of decreasing terms. I guess the point is that the difference between $\log\left(1+\frac{(-1)^{k+1}}{k^\alpha}\right)$ and $\frac{(-1)^{k+1}}{k^\alpha}$ is sufficiently large to make the sum of the former diverge. Commented Mar 20 at 12:20
• The name is "Leibniz", not "libnitz". Commented Mar 20 at 16:40

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 $$\alpha\le\frac12$$.

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 $$\alpha>\frac12$$.