Does $\sum\limits_{k=1}^\infty \sum\limits_{n=k}^\infty \frac{(-1)^{n+k}}{n}$ diverge? 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?
 A: 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.
A: Let $A_k=\sum_{n=k}^{\infty}\frac {(-1)^{n+k}}{n+k}.$ We have $$A_k=\sum_{m=0}^{\infty}B(m,k)$$ where $B(m,k) =\frac {1}{2k+2m}-\frac {1}{2k+2m+1}.$
We have $B(m,k)>B(m+1,k)>0 $. So $$A_k>\sum_{m=0}^{k-1}B(m,k)\geq$$ $$\geq  kB(k-1,k)=$$ $$=k(\frac {1}{4k-2}-\frac {1}{4k-1})=$$ $$=\frac {k}{(4k-2)(4k-1)}>$$ $$>\frac {k}{(4k)(4k)}=\frac {1}{16k}.$$ So $\sum_{k=1}^{\infty}A_k\geq \sum_{k=1}^{\infty}\frac {1}{16 k}=\infty.$
Note: $\frac {1}{16 k }$ is not meant to be an accurate estimate for $A_k.$ Just a useful lower bound.
