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How may I prove that $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}=\frac{\pi^4}{72}?$ I also discussed the problem in the chat, but no solution so far. Some hints? Thanks!

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    Your first one is incorrect, since \dfrac1{n(m^2+n^2)} > \dfrac1{n^2(m^2+n^2)}2012-12-23

2 Answers 2

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For now, here is how we can prove the second equality. Let the second sum be $S.$ Note that by symmetry we also have $S= \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2(m^2+n^2)}.$ Now adding the two forms gives: $2S = \sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{1}{m^2n^2}= \left( \sum_{m=1}^{\infty} \frac{1}{m^2} \right)\left( \sum_{n=1}^{\infty} \frac{1}{n^2}\right)= \frac{\pi^4}{36}.$

As Fabian alludes to in the comments, it appears the first equality does not hold, since the difference between the two sums is $\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2-n}{n^3}\frac{1}{(m^2+n^2)}>0.$

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    @Chris'ssister Yes you also have to interchange the sums, and for this it suffices to show the sum converges absolutely. I leave that as an exercise.2012-12-23
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I can derive the second half of your question. To do this, rewrite the double sum as

$\frac{1}{2} \sum_{n=1}^{\infty} \frac{1}{n^2} \left ( \sum_{m=-\infty}^{\infty} \frac{1}{m^2+n^2} - \frac{1}{n^2} \right )$

Use the fact that

$\sum_{m=-\infty}^{\infty} \frac{1}{m^2+n^2} = \frac{\pi}{n} \coth{\pi n}$

Now the sum is

$\frac{1}{2} \left ( \pi \sum_{n=1}^{\infty} \frac{\coth{\pi n}}{n^3} - \sum_{n=1}^{\infty}\frac{1}{n^4} \right )$

Now use the analysis here:

sum of series involving coth using complex analysis

to derive the following result:

$ \pi \sum_{n=1}^{\infty} \frac{\coth{\pi n}}{n^3} = \frac{7 \pi^4}{180} $

The result follows from the well-known result that $\sum_{n=1}^{\infty}\frac{1}{n^4} = \pi^4/90$.

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    Re-write the hyperbolic cotangent in terms of exponential functions and isolate $\zeta(3)$, you should obtain a fast converging series for Apery's constant atributed to Ramanujan, http://en.wikipedia.org/wiki/Ap%C3%A9ry's_constant, its the second series representation, this can also be restated in terms of the power series generating function of the divisor function namely, $\sum_{k=1}^\infty \frac{\sigma_3(n)}{n^3e^{2\pi k}}=\frac{7\pi^3}{360}-\frac{\zeta(3)}{2}$2013-01-15