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 \(p\) is a prime number, \(s\in R\) and \(s\gt 1\)

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

Note by Darkrai ~Rayquaza
1 month 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

\[\displaystyle \left(\sum_{n=1}^{\infty}Li_s\left(a^{\left(2n-1\right)}\right)\right)-\frac{1}{2^s}\left(\sum_{n=1}^{\infty}Li_s\left(a^{\left(4n-2\right)}\right)\right)\]

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

Aaghaz Mahajan - 1 month ago

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@Aaghaz Mahajan I don't get how you went to the real part of Polylogarithms but when I generalized the question, I saw that the numerator would always turn out to be \(\frac 12\) for all \(n \) except when \[n\equiv \lfloor p/2\rfloor \pmod p\]

Darkrai ~Rayquaza - 1 month ago

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Yes, I know......But, well I wanted to be rigorous instead of mere observation....:) But thanks anyways!!!

Aaghaz Mahajan - 1 month ago

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Should I send you my complete approach of getting to the expression above??

Aaghaz Mahajan - 1 month ago

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@Aaghaz Mahajan If it isn't much long and you have time, then you can post your approach....

Darkrai ~Rayquaza - 1 month ago

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Okay.....I'll do that tonight!! Btw, are you preparing for JEE?? Which class are you in?

Aaghaz Mahajan - 1 month ago

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@Aaghaz Mahajan No worries and Yes preparing for JEE, currently in class 12

Darkrai ~Rayquaza - 1 month ago

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