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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?

Olivier Oloa
  • 120,989
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    Useful note: $\frac{90}{\pi}$ is in degrees. In radians it is $\sin\left(\frac{1}{2}\right)$ – Winther Jun 29 '16 at 02:26
  • What method do you want to prove this with? If you really want to nuke the problem you could use an expansion for $\ln(x)$ valid in the neighbourhood of it's parameter (here it's about 1.0429) and then work from there. I also recommend looking up identities for the Bernoulli Numbers, and what defines them. When working out series expansions the definition will come up somewhere (though some substitutions may have to be made). While I am currently unable to attack the problem myself (not do I desire to without any work on your part first) the outlined technique might do it – Brevan Ellefsen Jun 29 '16 at 02:42
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    Integrate the Taylor-series for $\cot(x)$ and take $x=\frac{1}{2}$. – Winther Jun 29 '16 at 02:42

1 Answers1

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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.

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 $$ \sum_{k=0}^{\infty} B_k \frac{z^k}{k!} = \frac{z}{e^z-1}, \quad |z|<2\pi, $$ $B_0=1,\,B_1=-1/2, \,\ldots.$

Olivier Oloa
  • 120,989