Friday, Jan 13th

Problem
Sum the series $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}$

Solution

Exchanging $m$ and $n$ produces the same series, so the sum will be
$\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m(n3^m+m3^n)}+\frac{m n^2}{3^n(n3^m+m3^n)}=\frac{1}{2}\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{mn}{3^m 3^n}$

$=\frac{1}{2}\left(\sum_{k=1}^\infty \frac{k}{3^k}\right)^2=\frac{1}{2}\left(\frac{3}{4}\right)^2=\frac{9}{32}$