14

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!

Pranav Arora
  • 11,014

2 Answers2

20

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}

Start wearing purple
  • 53,234
  • 13
  • 164
  • 223
10

Hint: Expanding in terms of parial fractions:

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

David H
  • 29,921
  • 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
  • 2
    @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