# Double sum trouble

Evaluate: $$\sum_{j=1}^{\infty} \sum_{i=1}^{\infty} \frac{j^2i}{3^j(j3^i+i3^j)}$$

Honestly, I don't see where to start with this. I am sure that this is a trick question and I am missing something very obvious. I tried writing down a few terms for a fixed $j$ but I couldn't spot any pattern or some kind of easier series to handle.

Any help is appreciated. Thanks!

• I would suggest calculating first $\sum_{i=1}^{\infty} \frac{i}{(j3^i+i3^j)}$ – Test123 May 15 '14 at 16:46
• I have already tried that and that's actually the series I meant when I said "for a fixed $j$"....sorry, it was a poor choice of words. – Pranav Arora May 15 '14 at 16:48
• Incidentally, this is an old Putnam problem. – heropup May 16 '14 at 2:16

## 2 Answers

After symmetrization with respect to the exchange $i\leftrightarrow j$, the sum can be rewritten as \begin{align} \frac12\sum_{i,j=1}^{\infty} \left(\frac{j^2i}{3^j(j3^i+i3^j)}+\frac{i^2j}{3^i(j3^i+i3^j)}\right)=\frac12\sum_{i,j=1}^{\infty} \frac{i\cdot j}{3^i\cdot3^j}=\frac12\left(\sum_{i=1}^{\infty}\frac{i}{3^i}\right)^2=\frac{9}{32}. \end{align}

• Sorry for reviving this but I am curious as to why the comments I posted here are no more? :/ – Pranav Arora May 18 '14 at 19:54
• @PranavArora I have no idea, maybe moderator's work? Was there any question on my post? – Start wearing purple May 18 '14 at 19:58
• No, I don't think there was any question. There were two comments by me. I don't see why a moderator would delete the comments. :/ – Pranav Arora May 18 '14 at 20:06
• @PranavArora I had some of my comments deleted in the past. But these were of "thank you" type. – Start wearing purple May 18 '14 at 20:08
• Mine was the same (I mean "thank you" type), maybe its against the rules to thank the users. -_- – Pranav Arora May 18 '14 at 20:09

Hint: Expanding in terms of parial fractions:

$$\frac{1}{3^j (j 3^i + i 3^j)}=\frac{1}{j 3^i 3^j}-\frac{i}{j 3^i (i 3^j + j 3^i)}\\ \implies \frac{j^2i}{3^j (j 3^i + i 3^j)}=\frac{j i}{3^i 3^j}-\frac{j i^2}{3^i (i 3^j + j 3^i)}.$$

• This is good too but that partial fraction decomposition isn't very obvious, some motivation behind it? :) – Pranav Arora May 15 '14 at 17:03
• @PranavArora Would you consider the partial fraction decomposition of $\frac{1}{x(x+a)}=\frac{1}{ax}-\frac{1}{a(x+a)}$ pretty obvious? From there, just substitute $x=i\,3^j$ and $a=j\,3^i$. – David H May 15 '14 at 17:13
• Yep, thanks David H for your input! – Pranav Arora May 15 '14 at 17:14
• @PranavArora You're very welcome. Though after seeing O.L.'s symmetrization argument, I have to say I like that much better. – David H May 15 '14 at 17:20