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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 : $$ S = \sum\limits_{m = 1}^\infty {\sum\limits_{n = 1}^\infty {\frac{{m^2 n}}{{3^m (n \cdot 3^m + m \cdot 3^n )}}} } $$

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  • $\begingroup$ That's a really nice solution, why don't you like it?? It's definitely not presuming the answer and I doubt you can do better than that. $\endgroup$ – user2345215 Nov 9 '14 at 23:29
  • $\begingroup$ Where does that intuition come from? Why would one take that average? $\endgroup$ – gerald Nov 9 '14 at 23:35
  • $\begingroup$ It's a trick. You can exchange the sums and it doesn't change the value, so you add them up and get twice the original while it simplifies a lot. $\endgroup$ – user2345215 Nov 9 '14 at 23:36
  • $\begingroup$ The reason is that you can change the meaning of $m$ and $n$ and the sum stays the same. So why don't you use both definitions together by adding them up. Then to get your answer you need to divide by 2. $\endgroup$ – Alexander Vlasev Nov 9 '14 at 23:55
  • $\begingroup$ but where does the simplification from the second line come from? $\endgroup$ – gerald Nov 10 '14 at 0:29
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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.

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