A nice relationship between $\zeta$, $\pi$ and $e$ I just happened to see this equation today, any suggestions on how to prove it?
$$\sum_{n=1}^\infty{\frac{\zeta(2n)}{n(2n+1)4^n}}=\log{\frac{\pi}{e}}$$
 A: The left-hand side is $$\sum_{n\ge 1}\frac{1}{\Gamma (2n)n(2n+1)4^n}\int_0^\infty\frac{x^{2n-1} dx}{e^x-1}=\int_0^\infty\frac{dx}{e^x-1}\sum_{n\ge 1}\frac{(x/2)^{2n-1}}{(2n+1)!}\\=\int_0^\infty\frac{dx}{e^x-1}\frac{\sinh\tfrac{x}{2}-\tfrac{x}{2}}{(\tfrac{x}{2})^2}.$$The rest is an exercise in complex analysis.
A: An approach that does not require complex analysis:
$$
\sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^n}
=\sum_{n=1}^\infty \frac{\zeta(2n)}{n4^n}-2\sum_{n=1}^\infty \frac{\zeta(2n)}{(2n+1)4^n}=S_1-2S_2
$$
To calculate $S_1$:
$$\begin{align}
S_1=\sum_{n=1}^\infty \frac{\zeta(2n)}{n4^n}
&=\sum_{n=1}^\infty \sum_{k=1}^\infty \frac{1}{n(4k^2)^n}\\
&=\sum_{k=1}^\infty \sum_{n=1}^\infty \frac{1}{n(4k^2)^n}\\
&=-\sum_{k=1}^\infty \ln\bigg(1-\frac{1}{4k^2}\bigg)\\
&=-\ln \prod_{k=1}^\infty \frac{(2k+1)(2k-1)}{(2k)^2}\\
&=\ln(\pi/2) 
\end{align}$$
using the Wallis Product. 
You may calculate $S_2$ similarly.
A: 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):
$$\frac{1}{\Gamma\left(z\right)}=ze^{\gamma z}\prod_{k=1}^{\infty}\left(1+\frac{%
z}{k}\right)\mathrm{e}^{-z/k}\text{.}$$
A: Since 
\begin{align*}
\zeta(2n) &= \frac{(-1)^{n+1} B_{2n} (2\pi)^{2n}}{2(2n)!},
\end{align*}
for integers $n > 0$, the given sum $S$ is
\begin{align*}
S &= \sum_{n=1}^\infty \frac{\zeta(2n)}{n(2n+1)4^n}
    = \sum_{n=1}^\infty (-1)^{n+1} \frac{B_{2n}}{(2n)(2n+1)}\frac{\pi^{2n}}{(2n)!}
\end{align*}
But
\begin{align*}
f(z) &= \sum_{n=1}^\infty \frac{B_n}{n(n+1)} \frac{z^{n+1}}{n!}
\end{align*}
has $\operatorname{Im} f(\pi i) = \pi S$ and
\begin{align*}
f''(z) &= \frac{1}{z}\sum_{n=1}^\infty B_n \frac{z^n}{n!} = \frac{1}{z}\left(-1 + \frac{z}{e^z - 1}\right) = -\frac{1}{z} + \frac{1}{e^z - 1}
\end{align*}
by (a) definition of the Bernoulli numbers $B_n$. A bit of careful integration gives the required sum.
