6
$\begingroup$

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.

$\endgroup$
  • 1
    $\begingroup$ Where did you find this series? $\endgroup$ – Mhenni Benghorbal Nov 23 '13 at 22:54
12
$\begingroup$

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) $$ S = \frac{9}{32}. $$

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.