Q:

Proof of a Zeta function identity

How do I show that
$$
\sum_{n\geq 1} \frac{\zeta(2n)}{n(2n+1)}=\ln\frac{2\pi}{e}.
$$
I found this equation in my homework. I tried to integrate Zeta function's generating   function twice, but the result has Li function in it. Is there any simple method to prove it?

A:

You don't need to use complex analysis since
$$\log (\sin (x))=\log (x)-\sum _{n=1}^{\infty } \frac{ \zeta (2 n)\,x^{2 n}}{n \,\pi ^{2 n}}\tag{1}$$
can be integrated between $0$ and $\pi$, which in turn can be derived by integrating the expansion of $\cot x$ below:
$$ \cot (x)=\frac{1}{x}-2 x \sum _{n=1}^{\infty } \frac{1}{(n\,\pi)^2-x^2}$$
See for example: G. Boros and V. H. Moll, Irresistible Integrals, Cambridge: Cambridge University Press, 2004, page 10.
(1) can also be derived using the sine product formula.

