I want to find a closed form for the following infinite sum: $$\sum_{k=2}^{\infty} \frac{(-1)^k\cdot(k-1)}{k\cdot(k+1)}\cdot \zeta(k)$$ Is it possible? My approach was to transform it into a double series and then to rewrite it in terms of logarithms, but i didn't get something useful.



We have

$$ \sum_{k=2}^{\infty} \frac{(-1)^k\cdot(k-1)}{k\cdot(k+1)}\cdot \zeta(k) = 2 - \log (2\pi). \tag1 $$

Proof. Let $0<x<1$. Observe that $$ \sum_{k=2}^{\infty} \frac{(-1)^k}{k} x^k = x- \log(1+x) $$ gives $$ \sum_{k=2}^{\infty} \frac{(-1)^k\cdot(k-1)}{k\cdot(k+1)}\cdot x^k = -2+\left( \frac 2x +1\right) \log(1+x). \tag2 $$ Using $$ \zeta(k)=1+\sum_{p=2}^{\infty} \frac{1}{p^k}, \quad k\geq 2,$$ then interverting the two summations in the initial series leads to

$$ \sum_{k=2}^{\infty} \frac{(-1)^k\cdot(k-1)}{k\cdot(k+1)}\cdot \zeta(k) =\sum_{p=1}^{\infty} \left( -2+\left( 2p +1\right) \log\left(1+\frac1p\right)\right). \tag3 $$

For $N\geq 2$, we may rewrite the finite sum $$ \begin{align} \sum_{p=1}^{N} \left( -2+\left( 2p +1\right) \log\left(1+\frac1p\right)\right)& = \sum_{p=1}^{N} \left(\left(2p+1\right)\log\left(p+1\right)-\left(2p-1\right)\log p-2\log p-2\right) \\\\ & = \left(2N+1\right)\log\left(N+1\right)-2\log\left(N!\right)-2N \\\\ & =2 - \log (2\pi)+\mathcal{O}\left(\frac 1N\right) \end{align} $$ giving $(1)$ as $N$ tends to $+\infty$, where we have used Stirling's formula.

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    $\begingroup$ Thank you, this was what I mend with expressing it in terms of logarithms, but I wasn't able to do so. Very nice! $\endgroup$ – Redundant Aunt Sep 27 '14 at 10:57

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