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A zeta series

Let \zeta denote the Riemann zeta function. Evaluate the series

\displaystyle \mathcal{S} = \sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1) \left ( 2n+1 \right )}

(Seraphim Tsipelis, Anastasios Kotronis)

Solution

To what follows \zeta^* stands for

\zeta^*(n) = \left\{\begin{matrix} \zeta(n) & , & n \neq 1 \\ \gamma& , & n=1 \end{matrix}\right.

Then it is known that \displaystyle \sum_{n=1}^{\infty} \zeta^*(n) x^n = - x \psi^{(0)} (1-x) . Integrating we get that

(1)   \begin{equation*}\displaystyle \sum_{n=1}^{\infty} \frac{\zeta^*(n)}{n} x^n = \log \Gamma(1-x) \end{equation*}

Setting x \mapsto -x back at (1) and substracting these two equations we get that

(2)   \begin{equation*}\sum_{n=0}^{\infty} \frac{\zeta^*(2n+1)}{2n+1} x^{2n+1} = \frac{1}{2} \log \left ( \frac{\Gamma(1-x)}{\Gamma(1+x)} \right )\end{equation*}

Integrating both sides of (2) we get that

\begin{aligned} \frac{\gamma}{2} + \frac{1}{2} \sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1) (2n+1)} &= \frac{1}{2} \int_{0}^{1} \log \left ( \frac{\Gamma(1-x)}{\Gamma(1+x)} \right ) \, {\rm d}x \\ &= \frac{1}{2} \int_{0}^{1} \bigg [ \log \Gamma(1-x) - \\ & \quad \quad \quad \quad-\log \Gamma(1+x) \bigg ] \, {\rm d}x\\ &= \frac{1}{2} \end{aligned}

and thus

\displaystyle \sum_{n=1}^{\infty} \frac{\zeta(2n+1)}{(n+1)(2n+1)} = 1 - \gamma

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