TITLE: A nice relationship between $\zeta$, $\pi$ and $e$
QUESTION [5 upvotes]: I just happened to see this equation today, any suggestions on how to prove it?
$$\sum_{n=1}^\infty{\frac{\zeta(2n)}{n(2n+1)4^n}}=\log{\frac{\pi}{e}}$$

REPLY [2 votes]: The identities
$$\begin{align}\sum_{k=1}^{\infty}\frac{\zeta\left(2k\right)}{k}z^{2k}&=\ln\left(\frac{\pi z}{\sin\left(\pi z\right)}\right)&&(\lvert z \rvert < 1)\\
\sum_{k=1}^{\infty}\frac{\zeta\left(2k\right)}{(2k+1)2^{2k}}&=\frac{1}{2}-\frac
{1}{2}\ln 2\end{align}$$
(DLMF 25.8.8, 25.8.9) follow from the more basic identity
$$\begin{align}
\sum_{k=2}^{\infty}\frac{\zeta\left(k\right)}{k}z^{k}&=-\gamma z+\ln\Gamma\left
(1-z\right)&&(\lvert z \rvert < 1)
\end{align}$$
(DLMF 25.8.7), which in turn follows from the product formula for the Gamma function (DLMF 5.8.2):
$$\frac{1}{\Gamma\left(z\right)}=ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{%
z}{k}\right)\mathrm{e}^{-z/k}\text{.}$$