# Thread: Double Summation - Find sum of the series?

1. ## Double Summation - Find sum of the series?

Find sum of the series :

$\sum_{m = 1}^{\infty}\sum_{n = 1}^{\infty} \frac{m^{2}n}{3^{m}(n\cdot 3^{m} + m\cdot 3^{n})}$

2. since $\frac{m^{2}n}{3^{m}\left( n\cdot 3^{m}+m\cdot 3^{n} \right)}+\frac{mn^{2}}{3^{n}\left( m\cdot 3^{n}+n\cdot 3^{m} \right)}=\frac{mn}{3^{m+n}},$ your double sum equals $\frac{1}{2}\sum\limits_{m=1}^{\infty }{\sum\limits_{n=1}^{\infty }{\frac{mn}{3^{m+n}}}}=\frac{1}{2}\left( \sum\limits_{j=1}^{\infty }{\frac{j}{3^{j}}} \right)^{2}.$

now find that sum, which is not hard.