Putnam 1999, Problem A-4

Sum the series
\[
\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2 n}{3^m (n 3^m + m 3^n)}.
\]

Solution:
Denote the series by $S$, and let $a_n = 3^n/n$. Note that
$$S = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{a_m(a_m + a_n)}$$
$$= \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{a_n(a_m + a_n)},$$
where the second equality follows by interchanging $m$ and $n$. Thus
$$2S = \sum_{m,n} \left(\frac{1}{a_m(a_m + a_n)} + \frac{1}{a_n(a_m + a_n)}\right)$$
$$= \sum_{m,n} \frac{1}{a_ma_n}$$
$$= \left(\sum_{n=1}^\infty \frac{n}{3^n}\right)^2.$$

But
$$\sum_{n=1}^\infty \frac{n}{3^n} = \frac{3}{4}$$
since, e.g., it's $f'(1)$, where
$$f(x) = \sum_{n=0}^\infty \frac{x^n}{3^n} = \frac{3}{3-x},$$
and we conclude that $S = 9/32$.