I am a PhD student in electrical engineering. I need to find a closed form formula for the following series: $$\sum_{k=1}^{\infty}\frac{1}{2}A_k^2e^{-k^2\sigma_m^2}(e^{k^2\sigma_m^2}-1)$$where $A_k= \frac{4\sin(\frac{\pi}{2}k)}{\pi k}$ and $\sigma_m^2$ is a constant. This is a very important result if I can find it.

Thanks a lot.

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    $\begingroup$ This is unreadable, you should make the picture bigger, or just typeset it yourself. $\endgroup$ – Igor Rivin Dec 16 '13 at 3:31
  • $\begingroup$ my dear Trafalgar suggested an edited version of my formula. I hope it is clear now. Thanks so much, you guys are amazing $\endgroup$ – EE_engineer Dec 16 '13 at 3:51
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    $\begingroup$ On some webpage, I saw the formula $$\sum_{n=-\infty}^\infty e^{-\pi n^2} = \frac{\pi^{1/4}}{\Gamma(3/4)}.$$They asked how to prove the formula, and the reply was that it was in one of Ramanujan's notebooks. At this point everyone assumed it was either impossible to prove, or impossible to follow the argument. My guess is that your formula might be just as hard, or harder. $\endgroup$ – Stephen Montgomery-Smith Dec 16 '13 at 3:57
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    $\begingroup$ Do you know that this has a closed form? $\endgroup$ – Avi Steiner Dec 16 '13 at 4:12
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    $\begingroup$ Split this up into two sums: $$ \sum_{k=1}^\infty A_k - \sum_{k=1}^\infty A_k e^{-k^2\sigma_m^2}.$$ The first sum (according two Wolfram|Alpha) is 1. For the second sum, let $$f(z)=\sum_{k=1}^\infty \frac{4}{\pi k}\sin\left(\frac{\pi kz}{2}\right) e^{-k^2\sigma_m^2}.$$ Then $$ f'(z)=2\sum_{k=1}^\infty \cos\left(\frac{\pi kz}{2}\right) e^{-k^2\sigma_m^2}=\vartheta\left(\frac{\pi z}{2}; \,\sigma^2i\right)-1,$$ where $\vartheta$ is the Jacobi Theta Function. $\endgroup$ – Avi Steiner Dec 16 '13 at 4:54

First of all, the $\sin$ term is a red herring, since it is equal to $1$ for odd $k,$ and $0$ for even $k.$ Second, expanding the second term, you will get two sums. The first is

$$\sum_{k=1}^\infty \frac{8}{\pi k}^2 \sin^2(k\pi/2),$$ which is a multiple of the sum of inverses of odd squares, and is easy to evaluate.

The second sum is $$(8/\pi^2)\sum_{\mbox{odd $k$}} e^{-k^2/\sigma^2}/k^2.$$ Make the exponent $(-x^2 k^2/\sigma^2),$ and differentiate with respect to $x.$ You will get a sum of the form

$$a x\sum e^{-x^2 b k^2},$$ so a multiple of a linear function and a theta function, so you sum can be expressed via the integrals of such. I leave the rest up to you (since I have no idea what it is you are trying to get).

  • $\begingroup$ That hat fits your profile picture really well! :) $\endgroup$ – Prism Dec 16 '13 at 7:36
  • $\begingroup$ @IgorRivin, Thanks so much. I need to look more into this theta function, I still cannot figure out how the theta function for ax∑e−x2bk2 should be written. I appreciate if you can give more clues. $\endgroup$ – EE_engineer Dec 16 '13 at 8:37

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