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Calculation of $\displaystyle \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^2n}{3^n\left(m\cdot 3^n+n\cdot 3^m\right)}$.

$\bf{My\; Try::}$

Let $\displaystyle S = \sum_{m=1}^{\infty}\sum_{n=1}^{\infty}\frac{m^2n}{3^n\left(m\cdot 3^n+n\cdot 3^m\right)}=\sum_{m=1}^{\infty}\left[\frac{1}{3}\cdot \frac{m^2}{ \left(3m+3^m\right)}+\frac{1}{3^2}\cdot \frac{2m^2}{(3^2m+2\cdot 3^m)}+.........\right]$

Actually I did not understand how can i solve it, Help me

Thanks

juantheron
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  • Thanks for giving me a link , But can anyone explain me how can we write $\displaystyle\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}f(m,n) = \sum_{n=1}^{\infty}\sum_{m=1}^{\infty}f(n,m),$ Thanks – juantheron Nov 17 '14 at 02:53
  • There are certain conditions for that to hold true, do have a look here (http://www.math.ubc.ca/~feldman/m321/twosum.pdf). One can think of it as summing a rectangular array of elements: the sum across rows is the same as the sum across columns. – Yiyuan Lee Nov 17 '14 at 02:59

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