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# Zeta Power Sum

Prove that :

$\sum_{r=1}^{n-1} (2^{2r}-1) \zeta(2n-2r) \zeta(2r) = \dfrac{1}{2} (2^{2n} -1) \zeta(2n)$

Notation : $$\displaystyle \zeta(n) = \sum_{k=1}^{\infty} \dfrac{1}{k^n}$$ denotes the Riemann Zeta function.

This is a part of the set Formidable Series and Integrals

Note by Ishan Singh
1 year, 5 months ago

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