Calculate the sum of the double series: $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2n}{3^m(n3^m+m3^n)}$$
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4Replacing $m\leftrightarrow n;,$ and Then Add these two series. – juantheron Nov 22 '16 at 10:14
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@juantheron nice trick! it also applies if one replaces $3$ by some generic $a<1$. Also we can write $m^{l+1}n^l$ in the numerator for $l\in N$ – tired Nov 22 '16 at 10:19
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@juantheron Could you please clarify which are the two series? – vidyarthi Nov 22 '16 at 10:39
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@juantheron If you are going to post an answer I am ready to delete mine. – Robert Z Nov 22 '16 at 10:48
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http://math.stackexchange.com/questions/1025317/calculation-of-sum-m-1-infty-sum-n-1-infty-fracm2n3n-leftm-cdo – juantheron Nov 22 '16 at 10:51
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Hints.
As suggested by the very useful comment of juantheron (which deserves all the merit of this post), consider the two sums $$S_1=\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2n}{3^m(n3^m+m3^n)}\quad\mbox{and}\quad S_2=\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{n^2m}{3^n(m3^n+n3^m)}.$$ Then show:
i) $S_1=S_2$.
ii) $\displaystyle S_1+S_2=\sum_{m=1}^\infty \sum_{n=1}^\infty \left (\frac{m^2n}{3^m(n3^m+m3^n)}+ \frac{n^2m}{3^n(m3^n+n3^m)}\right)=\left(\sum_{n=1}^\infty\frac{n}{3^n}\right)\cdot \left(\sum_{m=1}^\infty \frac{m}{3^m}\right)$.
iii) Evaluate $\displaystyle \sum_{n=1}^\infty\frac{n}{3^n}$.
iv) Evaluate $S_1$.
Robert Z
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Can we interchange the order of summation because the series is convergent? – vidyarthi Nov 22 '16 at 10:42
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We can interchange the order of summation because we are summing positive terms. – Robert Z Nov 22 '16 at 10:49
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