Any hints on how to prove that $\ln{1\over 2\sin\left({90\over \pi}\right)}=\sum_{n=1}^{\infty}{(-1)^{n-1}B_{2n}\over 2n(2n)!}$?

Question

How do you prove that

$$ \ln\left(1 \over 2\sin\left(1/2\right)\right) = \sum_{n = 1}^{\infty}{\left(-1\right)^{n - 1}\,B_{2n} \over 2n\left(2n\right)!}\ ?\tag1 $$

where $B_{2n}$ is a Bernoulli number

Any hints?

Answer 1

Hint. One may recall that $$ \cot x = \frac1x+\sum_{n=1}^{\infty} B_{2n} \frac{(-1)^n 4^n x^{2n-1}}{(2n)!},\quad |x|<\frac{\pi}2. \tag1 $$ We are allowed to integrate the power series termwise obtaining $$ \log( \sin x)=\log x+\sum_{n=1}^{\infty} B_{2n} \frac{(-1)^n 4^n x^{2n}}{2n(2n)!},\quad |x|<\frac{\pi}2, \tag2 $$ then taking $x=\dfrac12$ gives the result.

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Remark. We may obtain $(1)$ from observing that the even function $(-2\pi,2\pi) \ni z \mapsto z\cot z$ rewrites $$ z\cot z=z\cdot \frac{\cos z}{\sin z}=iz\cdot\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}=iz\cdot\frac{e^{2iz}+1}{e^{2iz}-1}=iz+\frac{2iz}{e^{2iz}-1}$$ and from using the classic generating function of the Bernoulli numbers $$ \sum_{k=0}^{\infty} B_k \frac{z^k}{k!} = \frac{z}{e^z-1}, \quad |z|<2\pi, $$ $B_0=1,\,B_1=-1/2, \,\ldots.$