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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
11 months, 2 weeks ago

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