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.
A: Alternative solution:
For convenience, set $0\ln 0 = 0$. We have
\begin{align}
\sum_{n\ge 1} \frac{\zeta(2n)}{n(2n+1)} &= \sum_{n\ge 1} \frac{1}{n(2n+1)}\sum_{k\ge 1}\frac{1}{k^{2n}}\tag{1}\\
&= \sum_{k\ge 1} \sum_{n\ge 1} \frac{1}{n(2n+1)} \frac{1}{k^{2n}}\tag{2}\\
&= \sum_{k\ge 1}\Big((k-1)\ln (k-1) - (k+1)\ln (k+1) + 2\ln k + 2 \Big)\tag{3}\\
&= \lim_{n\to \infty} \left[ \sum_{k=1}^n \Big((k-1)\ln (k-1) - (k+1)\ln (k+1) + 2\ln k + 2 \Big)\right]\tag{4}\\
&= \lim_{n\to \infty}\Big( - n\ln n - (n+1)\ln (n+1) + 2\ln(n!) + 2n\Big)\tag{5} \\
&= \ln \frac{2\pi}{\mathrm{e}}.\tag{6}
\end{align}
Here, in $(2) \Rightarrow (3)$, we have used Fact 1 given later;
in $(5) \Rightarrow (6)$, we have used Stirling's formula
$$\ln n! = \ln \sqrt{2\pi n} + n\ln \frac{n}{\mathrm{e}} + \mathrm{O}\big(\frac{1}{n}\big).$$
$\phantom{2}$
Fact 1: Let $y > 1$. Then
$$\sum_{n\ge 1}\frac{1}{n(2n+1)}\frac{1}{y^{2n}}
= (y-1)\ln (y-1) - (y+1)\ln (y+1) + 2\ln y + 2.$$
Since the series in LHS converges at $y=1$ and the limit of RHS as $y$ approaches $1$ exists, from Abel's theorem,
the above formula works also for $y=1$.
Proof of Fact 1: First, we have
$$\sum_{n\ge 1} \frac{1}{n(2n+1)}\frac{1}{y^{2n}} = \sum_{n\ge 1} \frac{1}{n} \frac{1}{y^{2n}} - \sum_{n\ge 1} \frac{2}{2n+1} \frac{1}{y^{2n}}.$$
Second, we have
$$\sum_{n\ge 1} \frac{1}{n} \frac{1}{y^{2n}} = -\ln \Big(1 - \frac{1}{y^2}\Big).$$
Third, let
$$g(x) = \sum_{n\ge 1} \frac{2}{2n+1} \frac{1}{y^{2n}}x^{2n+1}, \quad x\in [0, 1].$$
We have
$$g'(x) = \sum_{n\ge 1} 2\frac{1}{y^{2n}}x^{2n} = \frac{2x^2}{y^2-x^2} =  -2 + \frac{y}{y-x} + \frac{y}{y+x}.$$
Since $g(0)=0$, we have
$$g(x) = -2x - y\ln (y-x) + y\ln (y+x)$$
which enables us to obtain (by letting $x=1$)
$$\sum_{n\ge 1} \frac{2}{2n+1} \frac{1}{y^{2n}} = -2 - y\ln (y-1) + y\ln (y+1).$$
The desired result follows. 
