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On my recent post I asked a similar question to (1) and was proven by Paramanand Singh and Marko Riedel see here

$\Gamma\left(\frac{3}{4}\right)=1.225416702...$

(1)

$$1+8\sum_{n=1}^{\infty}\frac{n}{e^{n\pi}+(-1)^n}=\frac{\pi}{\Gamma^4\left(\frac{3}{4}\right)}$$

(1) I am not so that is the closed form. Can anyone help us verify (1) please?

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This one is cool! I proved in this answer that $$\vartheta_{3}(q) = \frac{\sqrt[4]{\pi}}{\Gamma(3/4)}\tag{1}$$ if $q = e^{-\pi}$. And the sum in current question is $$F(q) = 1 + 8\sum_{n = 1}^{\infty}\frac{nq^{n}}{1 + (-1)^{n}q^{n}}\tag{2}$$ It is possible to show with some effort that $\vartheta_{3}^{4}(q) = F(q)$ and our job is done.

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  • That's an awesome site! Thanks for posting. – Noam Shalev - nospoon May 29 '16 at 10:52
  • @nospoon: I am glad that someone liked my blog. These days my activity at MSE is preventing me to add new stuff on the blog so you have to be content with the old stuff there (at least for now). – Paramanand Singh May 29 '16 at 11:04
  • Thank you @ParamanandSingh. Nice knowing you. Here at MSE meeting a lot of people with mountain of knowledge. I do a have another similar as above but yield a rational number. $\sum_{n=1}^{\infty}\frac{n}{(-1)^ne^{n\pi}+1}=-\frac{1}{24}$. Is this an another trivial type? Should I post this one also? –  May 29 '16 at 11:38
  • @mahdishafici: You can calculate this new series by expressing it as $$\sum(-1)^{n}\frac{nq^{n}}{1 + (-q)^{n}}$$ and further simplify it as sum/difference of series whose sum is easily calculated. BTW any of such series is non-trivial unless you know the link between such series and theta functions/elliptic integrals. You may post it if you wish. – Paramanand Singh May 29 '16 at 12:01
  • @mahdishafici: You can directly prove that your new sum is negative of the sum in question http://math.stackexchange.com/q/389146/72031 and hence it is $-1/24$. – Paramanand Singh May 29 '16 at 12:24