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!
Prove $\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n(m^2+n^2)}=\sum_{m=1}^{\infty}\sum_{n=1}^{\infty} \frac{1}{n^2(m^2+n^2)}=\frac{\pi^4}{72}$
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1Your first one is incorrect, since $\dfrac1{n(m^2+n^2)} > \dfrac1{n^2(m^2+n^2)}$ – 2012-12-23
2 Answers
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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1@Ragib Zaman: glad to see you back! :-) (+1) – 2012-12-23
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1I have the feeling that it is good that you are not able to show the first equality... – 2012-12-23
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0@Ragib Zaman: is the symmetry enough to justify that the 2 double series are equal? – 2012-12-23
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0@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 $\sum_{n=1}^{\infty} 1/n^2 = \sum_{m=1}^{\infty}1/m^2.$ – 2012-12-23
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0@ 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). – 2012-12-23
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0@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
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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0thanks for your solution! (+1) – 2012-12-23
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1Thanks for posting the problem! – 2012-12-23
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0Re-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