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

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

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}