Double alternating sum

Let $\displaystyle c_n= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k+n}$. Evaluate the sum:

$$\mathcal{S}=\sum_{n=1}^{\infty} \frac{c_n}{n}$$

Solution

Well we have successively:

\begin{align*}
\mathcal{S} &=\sum_{n=1}^{\infty} \frac{c_n}{n} \\
 &= \sum_{n=1}^{\infty} \frac{1}{n}\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k+n}\\
 &= \sum_{k=1}^{\infty} (-1)^{k-1} \sum_{n=1}^{\infty} \frac{1}{n\left ( k+n \right )}\\
 &=\sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \sum_{n=1}^{\infty} \left [ \frac{1}{n} - \frac{1}{k+n} \right ]  \\
 &= \sum_{k=1}^{\infty} \frac{(-1)^{k-1}}{k} \left ( \psi^{(0)} (k+1) + \gamma \right )\\
 &= \sum_{k=1}^{\infty} \frac{(-1)^{k-1} }{k} \left ( \mathcal{H}_k + \gamma - \gamma  \right ) \\
 &= \sum_{k=1}^{\infty} \frac{(-1)^{k-1} \mathcal{H}_k}{k} \\
 &= \frac{\pi^2}{12} - \frac{\log^2 2}{2} 
\end{align*}

since it is know that $\displaystyle \sum_{n=1}^{\infty} \mathcal{H}_n x^{n-1} = - \frac{\log(1-x)}{x \left ( 1-x \right )}$. Substituting $x \mapsto -x$ and integrating from $0$ to $1$we have the result.

Remarks: It is worth noting the following:

1. We can find a generating function that is $\displaystyle \sum_{n=1}^{\infty} c_n x^n$. The interested reader is welcome to try and share his thoughts.
2. Here is another interesting sum:
   
   $$\sum_{n=1}^{\infty} \frac{1}{n} \left ( \frac{1}{n+1} - \frac{1}{n+2} + \frac{1}{n+3} - \cdots \right )^2 = \sum_{n=1}^{\infty} \frac{c_n^2}{n} = \frac{\pi^2 \log 2}{6} - \frac{\log^3 2}{3} - \frac{3 \zeta(3)}{4}$$
   
   which is due to Ovidiu Furdui. This can be tackled using gen. functions and some other tricks. You are welcome to try it though.