A Tricky Double Sum

Question

I was trying to solve an integral, which ended up being a difficult double sum. $$ \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{k+n+1}}{k(k+2n)^2} $$ I tried to use symmetry, but in vain. The inner sum is similar to the alternating Hurwitz-Zeta function. Using it, I ended up with $$ \frac{1}{4}\sum_{k=1}^{\infty} \frac{(-1)^k}{k}\zeta^{*}\Big(2,\frac{k+2}{2}\Big) $$ How can I solve this sum?

Answer 1

$$S_k=\sum_{n=1}^{\infty} \frac{(-1)^{k+n+1}}{k(k+2n)^2}=\frac{(-1)^{k+1}}{16k}\left(\psi ^{(1)}\left(\frac{k+4}{4}\right)-\psi ^{(1)}\left(\frac{k+2}{4}\right)\right)$$ I do not know how to compute the outer sum but it seems to converge quite fast $$\left( \begin{array}{cc} p & \sum_{k=1}^{p} S_k \\ 10 & -0.0677739 \\ 20 & -0.0679219 \\ 30 & -0.0679398 \\ 40 & -0.0679445 \\ 50 & -0.0679462 \\ 60 & -0.0679470 \\ 70 & -0.0679474 \\ 80 & -0.0679476 \\ 90 & -0.0679477 \\ 100 & -0.0679478 \end{array} \right)$$

The result is in the form $$\sum_{k=1}^{p} S_k =a_p+b_p \,C + c_p\,\pi^2$$

Using the inverse symbolic calculator the result is close to $-\frac{11+10 \sqrt{11}}{65} $