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

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?

• Check out identity (58) in mathworld.wolfram.com/JacobiThetaFunctions.html – nospoon May 29 '16 at 10:35
• Thank you so much @nospoon, after all this year We though it was a new thing we found. – user339807 May 29 '16 at 11:28

This one is cool! I proved in this answer that $$\vartheta_{3}(q) = \frac{\sqrt{\pi}}{\Gamma(3/4)}\tag{1}$$ 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.
• 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? – user339807 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 math.stackexchange.com/q/389146/72031 and hence it is $-1/24$. – Paramanand Singh May 29 '16 at 12:24