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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!

user 1591719
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    Your first one is incorrect, since $\dfrac1{n(m^2+n^2)} > \dfrac1{n^2(m^2+n^2)}$ –  Dec 23 '12 at 09:50

2 Answers2

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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 $$\sum_{m=1}^{\infty} \sum_{n=1}^{\infty} \frac{n^2-n}{n^3}\frac{1}{(m^2+n^2)}>0.$$

Ragib Zaman
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    @Ragib Zaman: glad to see you back! :-) (+1) – user 1591719 Dec 23 '12 at 09:44
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    I have the feeling that it is good that you are not able to show the first equality... – Fabian Dec 23 '12 at 09:51
  • @Ragib Zaman: is the symmetry enough to justify that the 2 double series are equal? – user 1591719 Dec 23 '12 at 13:08
  • @Chris'ssister Sure. I have just noted that instead of using the letter $m$ I could have used $n$ and vice versa. Same reason why $\sum_{n=1}^{\infty} 1/n^2 = \sum_{m=1}^{\infty}1/m^2.$ – Ragib Zaman Dec 23 '12 at 16:36
  • @ Ragib Zaman: however, I'm afraid it's not the same thing without interchanging the sums, and the sums may be interchanged iff the double sum converges. What do you think? I agree with your example above, but in our case things are different (it seems to me). – user 1591719 Dec 23 '12 at 16:59
  • @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. – Ragib Zaman Dec 23 '12 at 23:42
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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 $\sum_{n=1}^{\infty}\frac{1}{n^4} = \pi^4/90$.

Community
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Ron Gordon
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  • thanks for your solution! (+1) – user 1591719 Dec 23 '12 at 09:56
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    Thanks for posting the problem! – Ron Gordon Dec 23 '12 at 09:59
  • 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, $$\sum_{k=1}^\infty \frac{\sigma_3(n)}{n^3e^{2\pi k}}=\frac{7\pi^3}{360}-\frac{\zeta(3)}{2}$$ – Ethan Splaver Jan 15 '13 at 23:58