What is the solution to Problem A-4 in the 1999 Putnam Mathematical Competition?

[Ackbach]

Here is this week's POTW:

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

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[Ackbach]

Re: Problem Of The Week # 251 - Feb 02, 2017

This was Problem A-4 in the 1999 William Lowell Putnam Mathematical Competition.

Trying to get back to an earlier day in the week for the POTW schedule, so I'm going to go ahead and close this one. No one answered this week's POTW. The solution, attributed to Kiran Kedlaya and his associates, follows:

Spoiler

Denote the series by $S$, and let $a_n = 3^n/n$. Note that
\begin{align*}
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)},
\end{align*}
where the second equality follows by interchanging $m$ and $n$. Thus
\begin{align*}
2S &= \sum_m \sum_n \left( \frac{1}{a_m(a_m+a_n)} +
\frac{1}{a_n(a_m+a_n)}\right) \\
&= \sum_m \sum_n \frac{1}{a_m a_n} \\
&= \left( \sum_{n=1}^\infty \frac{n}{3^n} \right)^2.
\end{align*}
But
\[
\sum_{n=1}^\infty \frac{n}{3^n} = \frac34
\]
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 $S = 9/32$.