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I'm studying for the Putnam Exam and am a bit confused about how to go about solving this problem.

Sum the series $$ \sum_{m = 1}^{\infty} \sum_{n = 1}^{\infty} \frac{m^2n}{3^m(n3^m + m3^n)}. $$

I've tried "splitting" the expression to see if a geometric sum pops up but that didn't get me anywhere. I've also tried examining the first few terms of the series for the first few values of $m$ to see if an inductive pattern emerged but no luck there either.

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Where did you find this series? –  Mhenni Benghorbal Nov 23 '13 at 22:54

1 Answer 1

Let $$ S = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{m^2n}{3^m(n3^m + m3^n)}= \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{\frac{3^m}{m}\left( \frac{3^m}{m} + \frac{3^n}{n} \right)}. $$ Then we see by symmetry, that $$ 2S = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{\left( \frac{3^m}{m} + \frac{3^n}{n} \right)}\left( \frac{1}{\frac{3^m}{m}} + \frac{1}{\frac{3^n}{n}}\right), $$ or what is the same, $$ 2S = \sum_{m=1}^\infty\sum_{n=1}^\infty \frac{mn}{3^m 3^n} = \left( \sum_{m=1}^\infty \frac{m}{3^m} \right)^2. $$ The problem reduces to calculating the last single sum.

To this end, recall that for $|x|< 1,$ geometric series yields $$ \frac{1}{1-x} = 1 + x+ x^2 + \cdots, $$ multiplying by $x,$ and differentiating (this is justified because the series on the right converges on compact subsets of $|x| < 1,$ $$ \frac{1}{(1-x)^2} = 1+ 2x + 3x^2 + \cdots, $$ and multiplying by $x$ one more time, $$ \frac{x}{(1-x)^2} = x + 2x^2 + 3x^3 + \cdots. $$ Set $x = 1/3$ to evaluate the single sum on the right. We obtain (if I haven't messed up calculations) $$ S = \frac{9}{32}. $$

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(+1) nice technique. –  Mhenni Benghorbal Nov 23 '13 at 23:38
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Thanks very much :) –  Raghav Nov 23 '13 at 23:43

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