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

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.

• Hint: Assume $f(j,k) = f(k,j)$, we have $\sum\limits_{j=1}^\infty\sum\limits_{k=1}^\infty f(j,j+k) = \sum\limits_{1\le j < k} f(j,k) = \frac12 \sum\limits_{1 \le j \ne k} f(j,k) = \frac12 \left[\sum\limits_{1\le j,k}f(j,k) - \sum\limits_{1\le j}f(j,j)\right]$ – achille hui Sep 29 '18 at 0:37

## 1 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}

• Are you able to explain why $\sum^\infty_{j=1}\sum^\infty_{n=j+1}\frac{1}{j^2n^2} = \frac{1}{2}\left(\sum^\infty_{j=1}\sum^\infty_{n=1} \frac{1}{j^2n^2}-\sum^\infty_{j=1} \frac{1}{j^4}\right)\\$ – Tom Himler Sep 29 '18 at 2:38
• @TomHimler If you make a lattice diagram, you will see that the sum is over half the lattice diagram. – Jacky Chong Sep 29 '18 at 2:39
• Sorry, I'm not sure what a lattice diagram is. – Tom Himler Sep 29 '18 at 2:48
• plot the points $(j, n) = (1, 2), (1, 3) (1, 4)...$. This is what I called the lattice diagram. – Jacky Chong Sep 29 '18 at 2:55
• Sorry, something is still not clicking. I understand the diagram now, but I'm not sure how the result follows from that. – Tom Himler Sep 29 '18 at 3:13