How would I go about solving $\sum_{j=1}^\infty\sum_{k=1}^\infty \frac{1}{j^2(j+k)^2}$?

Question

I've got this sum that I've contrived. Wolfram has been able to evaluate it to $\frac{\pi^4}{120}$ but I'm not sure how to get there. I've tried switching j and k adding the sum together then averaging but nothing seems to come out it. I have no idea what else to try to do. I definitely feel like it should be in terms of $\zeta(4)$ but I'm not sure how. I have not dealt with double infinite series before.

Answer 1

Following @achillehui's hint, we see that \begin{align} \sum^\infty_{j=1}\sum^\infty_{k=1} \frac{1}{j^2(j+k)^2} =&\ \sum^\infty_{j=1}\sum^\infty_{n=j+1}\frac{1}{j^2n^2} = \frac{1}{2}\left(\sum^\infty_{j=1}\sum^\infty_{n=1} \frac{1}{j^2n^2}-\sum^\infty_{j=1} \frac{1}{j^4}\right)\\ =&\ \frac{1}{2}\left(\zeta(2)^2-\zeta(4) \right) = \frac{1}{2}\left(\frac{\pi^4}{36}-\frac{\pi^4}{90}\right)= \frac{\pi^4}{120}. \end{align}