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Math Help - Double Summation - Find sum of the series?

  1. #1
    Super Member fardeen_gen's Avatar
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    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})}
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  2. #2
    Math Engineering Student
    Krizalid's Avatar
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    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.
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