Interesting Trig sum

Prove/Disprove that

\[\displaystyle \sum_{n=0}^{\infty} \left(\frac {\displaystyle\sum_{k=0}^{\left \lfloor \frac p2 \right\rfloor -1} \left(\cos \left(\frac {\pi}{p}(2k+1)(2n+1)\right)\right) }{(2n+1)^s}\right) =\left(\frac 12 -\frac {1}{2p^s }-\frac {\left\lfloor \frac p2\right\rfloor}{p^s}\right) (1-2^{-s})\zeta(s)\]

Where pp is a prime number, sRs\in R and s>1s\gt 1

Also .\lfloor. \rfloor denotes the floor function while ζ(s)\zeta(s) denotes the Riemann Zeta function.

Note by Rohan Shinde
2 years, 4 months ago

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@Darkrai ~Rayquaza Any references???? I converted the sum into real parts of Polylogs but got stuck in finding the infinite summation.........

Basically, after a lot of simplifications, the problem breaks down to finding the real part of

(n=1Lis(a(2n1)))12s(n=1Lis(a(4n2)))\displaystyle \left(\sum_{n=1}^{\infty}Li_s\left(a^{\left(2n-1\right)}\right)\right)-\frac{1}{2^s}\left(\sum_{n=1}^{\infty}Li_s\left(a^{\left(4n-2\right)}\right)\right)

Here, aa is eiπp\displaystyle e^{\frac{i\pi}{p}} and Lis(a)\displaystyle Li_s\left(a\right) denotes the Polylogarithm of aa to the base ss

Aaghaz Mahajan - 2 years, 4 months ago

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