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

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.
