# 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?

## 2 Answers

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.

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.