Computing the value of a double series This is the first double series I've ever seen, and I am clueless on how to solve this. While looking for more theoretical aspects of double series, I accidentally found the solution here.
I don't like the listed solution because it seems really counterintuitive to me, it feels like  one of those solutions one got by reverse walking from an already known answer.
Thus being said, here's the problem at hand : $$
S = \sum\limits_{m = 1}^\infty  {\sum\limits_{n = 1}^\infty  {\frac{{m^2 n}}{{3^m (n \cdot 3^m  + m \cdot 3^n )}}} } $$
 A: The goal is to evaluate the sum of the double infinite series,
$$S=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}a_{m,n},$$
where the terms $a_{m,n}$ are given by the two-variable function,
$$a_{m,n}=\frac{m^2n}{3^m\left(n\cdot3^m+m\cdot3^n\right)}.$$
It will probably be much easier to see how one might exploit the symmetry if we reorganize the expression for $a_{m,n}$ a little. If we introduce a bit of auxiliary notation and define the single-variable function, $c_{x}:=x\cdot 3^{-x}$, we find a nice compact representation of $a_{m,n}$ in terms of this single function that is much more suggestive of the "near-symmetry" in $m$ and $n$:
$$\begin{align}
a_{m,n}
&=\frac{m^2n}{3^m\left(n\cdot3^m+m\cdot3^n\right)}\\
&=\left(m\cdot 3^{-m}\right)\frac{mn}{\left(n\cdot3^m+m\cdot3^n\right)}\\
&=\left(m\cdot 3^{-m}\right)\frac{m\left(n\cdot 3^{-n}\right)}{\left(n\cdot 3^{-n}\cdot3^m+m\right)}\\
&=\left(m\cdot 3^{-m}\right)\frac{\left(m\cdot 3^{-m}\right)\left(n\cdot 3^{-n}\right)}{\left(n\cdot 3^{-n}+m\cdot 3^{-m}\right)}\\
&=c_{m}\cdot\frac{c_{m}c_{n}}{c_{m}+c_{n}}.\\
\end{align}$$
Now when we take the symmetric sum of the elements it's immediately apparent that the double summation is now separable:
$$a_{m,n}+a_{n,m}=c_{m}\cdot\frac{c_{m}c_{n}}{c_{m}+c_{n}}+c_{n}\cdot\frac{c_{m}c_{n}}{c_{m}+c_{n}}=\left(c_{m}+c_{n}\right)\cdot\frac{c_{m}c_{n}}{c_{m}+c_{n}}=c_{m}c_{n}\\
\implies \sum_{m=0}^{\infty}\sum_{n=0}^{\infty}\left(a_{m,n}+a_{n,m}\right)=\sum_{m=0}^{\infty}\sum_{n=0}^{\infty}c_{m}c_{n}=\left(\sum_{m=0}^{\infty}c_{m}\right)\cdot\left(\sum_{n=0}^{\infty}c_{n}\right)=\left(\sum_{m=0}^{\infty}c_{m}\right)^2.$$
Hopefully that clears things up somewhat, but if not then please feel free to ask for further details.
