Infinite sum containing the Bernoulli numbers $B_{2n}$ I am looking for a more direct way to show that

$$\sum_{n = 0}^\infty \frac{(-1)^n B_{2n} \pi^{2n}}{(2n + 1)!} = \ln (2)$$

Here $B_n$ denote the Bernoulli numbers.
By more direct I mean a proof that starts preferably with the sum, then uses it in some way, before arriving at the result of $\ln (2)$.
I have managed to prove it in a very indirect way as follows. Starting with the well-known result of (see here for example)
$$\int_0^{\pi/2} x \cot x \, dx = \frac{\pi}{2} \ln (2),$$
using the Maclaurin series expansion for the cotangent function of
$$\cot x = \sum_{n = 0}^\infty \frac{(-1)^n 2^{2n} B_{2n}}{(2n)!} x^{2n - 1}, \quad 0 < |x| < \pi,$$
on substituting this result into the above integral, after interchanging the summation with the integral, and performing the resulting integration, one arrives at the result in question and is essentially the same method that was used here. But does a more direct way exist?  
 A: Since $B_0 = 1$, $B_1 = -1/2$ and $B_{2n+1}= 0$ for $n \geqslant 1$, using the generating function we have
$$\frac{t}{e^{t} - 1} = 1 - \frac{t}{2} +\sum_{n=1}^\infty \frac{B_{2n}t^{2n}}{(2n)!},$$
and
$$\frac{it}{e^{it} - 1} = 1 - \frac{it}{2}+\sum_{n=1}^\infty \frac{(-1)^nB_{2n}t^{2n}}{(2n)!}.$$
Integration over $[0,\pi]$ and some rearrangement should produce what you want.
Note that
$$\int_0^\pi \frac{it}{e^{it}-1}\, dt = \pi\log 2  - \frac{i \pi^2}{4}$$
A: Your approach looks pretty direct to me.
Starting from the generating function for the Bernoulli numbers 
$$\cot x = \sum_{n = 0}^\infty \frac{(-1)^n 2^{2n} B_{2n}}{(2n)!} x^{2n - 1},\qquad 0<|x|<\pi $$
is a straightfoward consequence, and the relation between the wanted series and $\int_{0}^{\pi/2}x\cot x\,dx$ is clear as well. By integration by parts the latter integral only depends on $\int_{0}^{\pi/2}\log\sin x\,dx$, which on its turn can be computed from symmetry or Riemann sums, via
$$ \prod_{k=1}^{n-1}\sin\left(\frac{\pi k}{n}\right) = \frac{2n}{\color{red}{2}^n}.$$
Ultimately, $\log 2$ arises from the highlighted $2$.
