A series

[Tolaso J Kos]

Evaluate the series:

$$\mathcal{S}=\sum_{n=0}^{\infty} \left ( \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{n+k} \right )^2$$
(Limits, Series and Fractional Part Integrals: Problems in Mathematical Analysis)

[r9m]

\begin{align*}\sum_{n=0}^{\infty} \left(\sum_{k=1}^{\infty} \frac{(-1)^{k}}{n+k}\right)^2 &= \sum\limits_{n=0}^{\infty} \left((-1)^{n}\int_0^{1} \frac{x^n}{1+x}\,dx\right)^2\\&= \sum\limits_{n=0}^{\infty} \int_0^{1}\int_0^1 \frac{x^ny^n}{(1+x)(1+y)}\,dx\, dy\\&= \int_0^{1} \int_0^1 \frac{\,dx \,dy}{(1-xy)(1+x)(1+y)}\\&= \int_0^1 \left(\int_0^1 \frac{y}{1-xy} + \frac{1}{1+x}\,dx\right)\frac{\,dy}{(1+y)^2}\\&= \int_0^1 \frac{\log 2 - \log (1-y)}{(1+y)^2}\,dy\\&= -\frac{1}{4}\int_0^1 \frac{\log (y/2)}{\left(1-\frac{y}{2}\right)^2}\,dy\\&= -\frac{1}{2}\int_0^{1/2} \frac{\log y}{(1-y)^2}\,dy\\&= -\frac{1}{2}\int_0^{1/2} \sum\limits_{n=1}^{\infty} ny^{n-1}\log y\,dy\\&= \frac{1}{2}\sum\limits_{n=1}^{\infty} \frac{1+n\log 2}{n2^n} = \log 2\end{align*}