How to evaluate the following infinite sum involving the Riemann zeta function? 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.
Thanks!
 A: 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.
