Does $\displaystyle\sum_{k=1}^\infty \sum_{n=k}^\infty \frac{(-1)^{n+k}}{n}$ diverge?
It is clear that the alternating Harmonic series converges:
$$\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n}=\log 2.$$
Thus, $S_k=\displaystyle\sum_{n=k}^\infty \frac{(-1)^{n+k}}{n}$ converges for each $k$. For each $k$, the sum could be expressed either as $\log 2-\alpha$ or $\alpha -\log 2$. So, we're really only interested in how much $S_k$ deviates $(\alpha)$ from the alternating series $S_1$. The numerators of $S_k$ follow for even $k$ and for odd $k$.
However, it seems that the partial sums may slowly go towards infinity as shown in this Wolfram plot here. Reasonably, since the difference between $S_1$ and $S_k$ probably behaves like $O(1/k)$, the sum probably diverges.
What would be the best way to show convergence/divergence?
Best Answer
The sum of the first two terms is $$ \begin{align} &\phantom{=1}1-\frac12+\frac13-\frac14+\frac15-\frac16+\dots\\ &\phantom{=1-1}\frac12-\frac13+\frac14-\frac15+\frac16-\dots\\ &=1 \end{align} $$ The sum of the next two terms is $$ \begin{align} &\phantom{=\frac13}\frac13-\frac14+\frac15-\frac16+\dots\\ &\phantom{=\frac13-\frac13}\frac14-\frac15+\frac16-\dots\\ &\phantom{1}=\frac13 \end{align} $$ The sum of the first $2n$ terms is $$ H_{2n}-\frac12H_n\sim\frac12\log(n)+\log(2)+\frac12\gamma $$ Thus, the series does diverge.