Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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!

share|improve this question
I would suggest calculating first $\sum_{i=1}^{\infty} \frac{i}{(j3^i+i3^j)}$ –  Test123 May 15 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 at 16:48
Incidentally, this is an old Putnam problem. –  heropup May 16 at 2:16

2 Answers 2

up vote 20 down vote accepted

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}

share|improve this answer
Sorry for reviving this but I am curious as to why the comments I posted here are no more? :/ –  Pranav Arora May 18 at 19:54
@PranavArora I have no idea, maybe moderator's work? Was there any question on my post? –  O.L. May 18 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 at 20:06
@PranavArora I had some of my comments deleted in the past. But these were of "thank you" type. –  O.L. May 18 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 at 20:09

Hint: Expanding in terms of parial fractions:

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

share|improve this answer
This is good too but that partial fraction decomposition isn't very obvious, some motivation behind it? :) –  Pranav Arora May 15 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 at 17:13
Yep, thanks David H for your input! –  Pranav Arora May 15 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 at 17:20

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.