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To find the value of $$\sum_{m=1}^{∞}{\sum_{n=1}^{∞}{\frac{m^2\cdot n}{3^m \cdot (n\cdot 3^m+m\cdot3^n)} } }$$ I dont know how to proceed to these kind of problems. Can anybody provide a sol to this problem which may give me an insight to solve more like these :)

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  • $\begingroup$ Are you confortable with solving ${\sum_{n=1}^{∞}{\frac{An}{Bn+C\cdot3^n} } } $? $\endgroup$ – Martigan Sep 18 '14 at 15:17
  • $\begingroup$ Actually, no, But if u can provide some link or some kind off, I will try to understand $\endgroup$ – Dinesh Sep 18 '14 at 15:19
  • $\begingroup$ @Martigan: I think no one is comfortable with solving that :D $\endgroup$ – Jack D'Aurizio Sep 18 '14 at 15:25
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    $\begingroup$ @JackD'Aurizio You are right... :D... $\endgroup$ – Martigan Sep 18 '14 at 15:35
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We have, with a classical symmetry trick: $$2\sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\frac{m^2 n}{3^m(m 3^n + n 3^m)}=\sum_{n=1}^{+\infty}\sum_{n=1}^{+\infty}\frac{\frac{m^2n}{3^m}+\frac{mn^2}{3^n}}{m3^n+n3^m}=\sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\frac{mn}{3^{m+n}}=\left(\sum_{n=1}^{+\infty}\frac{n}{3^n}\right)^2$$ hence the original series just equals $\frac{1}{2}\left(\frac{3}{4}\right)^2 = \color{red}{\frac{9}{32}}.$

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    $\begingroup$ I was expecting that from you ! Bravo ! $\endgroup$ – Claude Leibovici Sep 18 '14 at 15:30
  • $\begingroup$ what's the trick, please explain and how did you transformed second into third form (where n and m are interchanged) $\endgroup$ – RE60K Sep 18 '14 at 16:14
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    $\begingroup$ Just switch the role of $m$ and $n$. We have:$$\sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\frac{m^2n}{3^m(m3^n+n3^m)} = \sum_{n=1}^{+\infty}\sum_{m=1}^{+\infty}\frac{mn^2}{3^n(m3^n+n3^m)}.$$ Then sum and simplify. $\endgroup$ – Jack D'Aurizio Sep 18 '14 at 17:13
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    $\begingroup$ Fantastic as always !!! (+1) :D $\endgroup$ – r9m Nov 7 '14 at 16:32

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