Alternating series involving zeta function Can anyone help me attain the result for the following series?
$$\sum_{n=2}^{\infty} \frac{(-1)^n \zeta(n)}{n(n+1)}= \frac{1}{2} \left( \log 2 + \log \pi +\gamma -2 \right)$$
I don't know how to start. I am seriously thinking that this can be done using residues or contour integration since with real analysis I cannot see a pattern. 
 A: Take the logarithm of the Weierstass product form of the Gamma function 
$$ \frac{1}{\Gamma(x)}=xe^{\gamma x}\prod_{n=1}^\infty \left( 1+\frac{x}{n}\right)e^{-x/n}$$
to obtain $$-\log\Gamma(x)=\log x+\gamma x+\sum_{n=1}^\infty \log\left(1+\frac{x}{n}\right)-\frac{x}{n}$$
Now, $$\sum_{n=1}^\infty \log\left(1+\frac{x}{n}\right)-\frac{x}{n} =-\sum_{n=1}^\infty\sum_{m=2}^\infty\frac{\left(-\frac{x}{n}\right)^m}{m}
\\\\=-\sum_{m=2}^\infty\frac{(-1)^m x^m}{m}\sum_{n=1}^\infty\frac{1}{n^m}=-\sum_{m=2}^\infty\frac{(-1)^m x^m \zeta(m)}{m}$$
so that $$\sum_{n=2}^\infty\frac{(-1)^n  \zeta(n)}{n}x^n=\gamma x+\log(x)+\log\Gamma(x)$$
and $$\sum_{n=2}^\infty\frac{(-1)^n \zeta(n)}{n(n+1)}=\int_{0}^{1}(\gamma x+\log(x)+\log\Gamma(x))dx=\frac{\gamma}{2}-1+\int_{0}^{1}\log\Gamma(x)dx$$
 $\int_{0}^{1}\log\Gamma(x)dx=\log\sqrt{2\pi}$ is quite famous.
check here for reduction to evaluating $\int_0^\pi \log\sin(x)dx$
and here for the evaluation
A: it it not difficult to show that
$$\sum _{k=2}^{\infty } \frac{\pi  (-x)^k \zeta (k)}{k (k+1)}=\int_0^{\infty } -\frac{(t-2 i \pi  x) \log \left(1+\frac{i t}{2 \pi  x}\right)+(t+2 i \pi  x) \log \left(1-\frac{i t}{2 \pi  x}\right)-2 t}{\left(e^t-1\right) (4 \pi  x)} \, dt+\frac{1}{2} (-\pi  \log (x)+\pi -\pi  \log (2 \pi ))+\frac{1}{4} (-2 \gamma  \pi  x+3 \pi  x-2 \pi  x \log (x))$$
