Closed form for $\sum_{n=1}^{\infty} \sum_{k=1}^{n-1} \frac{(-1)^n}{k(n-k)}$

Question

Mathematica gives $$\sum_{n=1}^{\infty} \sum_{k=1}^{n-1} \frac{(-1)^n}{k(n-k)}=0.480453...$$ The question is: Can one get a closed form for this summation by hand?

Answer 1

Simply interchange the order of summation. Indeed, we have \begin{align} \sum_{n=1}^\infty \sum_{k=1}^{n-1} \frac{(-1)^n}{k(n-k)} &= \sum_{k=1}^\infty \sum_{n=k+1}^\infty \frac{(-1)^n}{k(n-k)} \\ &= \sum_{k=1}^\infty \sum_{m=1}^\infty \frac{(-1)^{k+m}}{km}\\ &= \sum_{k=1}^\infty \frac{(-1)^k}{k} \sum_{m=1}^\infty \frac{(-1)^m}{m}\\ &= \left(\sum_{k=1}^\infty \frac{(-1)^k}{k}\right)^2\\ &= (\ln 2)^2. \end{align}

Answer 2

With

$$ \frac1{k(n-k)}=\frac1n\left(\frac1k+\frac1{n-k}\right) $$

we can rewrite the sum as

\begin{eqnarray} \sum_{n=2}^\infty\sum_{k=1}^{n-1}\frac{(-1)^n}{k(n-k)} &=& \sum_{n=2}^\infty\frac{(-1)^n}n\sum_{k=1}^{n-1}\left(\frac1k+\frac1{n-k}\right) \\ &=& 2\sum_{n=2}^\infty(-1)^n\frac{H_{n-1}}n\;, \\ &=& 2\sum_{n=2}^\infty(-1)^n\frac{H_n}n-2\sum_{n=2}^\infty(-1)^n\frac1{n^2} \\ &=& 2\sum_{n=1}^\infty(-1)^n\frac{H_n}n+\frac{\pi^2}6\;, \end{eqnarray}

where $H_k$ is the $k$-th harmonic number.

From this answer by Sangchul Lee to Infinite Series $\sum\limits_{n=1}^\infty\left(\frac{H_n}n\right)^2$ we have

$$ \sum_{n=1}^\infty\frac{H_n}nz^n=-\operatorname{Li}_2\left(\frac z{z-1}\right)\;, $$

and substituting $z=-1$ yields

\begin{eqnarray} \sum_{n=1}^\infty(-1)^n\frac{H_n}n &=& -\operatorname{Li}_2\left(\frac12\right) \\ &=& \frac12\left(\log^22-\frac{\pi^2}6\right)\;. \end{eqnarray}

Thus your sum evaluates to $\log^22$, in agreement with the numerical result.

With hindsight, we should have recognized that your sum is just the square of the series

$$ \log2=-\sum_{n=1}^\infty(-1)^n\frac1n\;, $$

computed as the Cauchy product.