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A Bernoulli series

Let \mathcal{B}_n denote the n-th Bernoulli number. Prove that

    \[\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{B}_{2n} x^{2n}}{2n \left ( 2n \right )!} = \log \frac{x}{2} - \log \sin \frac{x}{2}\]

Solution

We begin by the known fact that

    \[\mathcal{B}_0 =0 \; , \; \mathcal{B}_1 = - \frac{1}{2} \; , \; \mathcal{B}_{2n+1} =0 \;\; \text{forall} \;\; n =1, 2, \dots\]

It is also well known that the generating function of the Bernoulli numbers is

    \[\sum_{n=0}^{\infty} \frac{\mathcal{B}_n x^n}{n!} = \frac{x}{e^x-1}\]

A simple observation shows that

    \begin{align*} \frac{x}{e^x-1} &= \sum_{n=0}^{\infty} \frac{\mathcal{B}_n x^n}{n!} \\ &=\sum_{n=0}^{\infty} \frac{\mathcal{B}_{2n} x^{2n}}{(2n)!} + \sum_{n=0}^{\infty} \frac{\mathcal{B}_{2n+1} x^{2n+1}}{\left ( 2n+1 \right )!} \\ &= 1 - \frac{x}{2} + \sum_{n=1}^{\infty} \frac{\mathcal{B}_{2n} x^{2n}}{\left ( 2n \right )!} + \\ & \quad \quad + \cancelto{0}{\sum_{n=1}^{\infty} \frac{\mathcal{B}_{2n+1} x^{2n+1}}{\left ( 2n+1 \right )!}} \\ &= 1 - \frac{x}{2} + \sum_{n=1}^{\infty} \frac{\mathcal{B}_{2n} x^{2n}}{\left ( 2n \right )!} \end{align*}

We have thus establised the identity

(1)   \begin{equation*} \sum_{n=1}^{\infty} \frac{\mathcal{B}_{2n} x^{2n}}{\left ( 2n \right )!} = \frac{x}{e^x-1} - 1 + \frac{x}{2} \end{equation*}

Setting x \mapsto it we get that

(2)   \begin{equation*}\sum_{n=1}^{\infty} \frac{(-1)^{n-1} \mathcal{B}_{2n} t^{2n-1}}{\left ( 2n \right )!} = \frac{1}{t} - \frac{i}{2} -\frac{i}{e^{it} -1 }\end{equation*}

Integrating (2) from \epsilon to x and taking limit as \epsilon \rightarrow 0 we get the result.

By the way, it also holds that

    \[\sum_{n = 0}^\infty {\frac{{{{\left( { - 1} \right)}^n}{2^{2n}}{\mathcal{B}_{2n}}{z^{2n - 1}}}}{{\left( {2n} \right)!}}} = \frac{1}{{\tan z}}\;,\;\left| z \right| < \pi\]

however the above series is not elementary.

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