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Let us consider an expansion $z \cot(z) = \sum_{n=0}^{\infty}{(-4)^{n} \cdot B_{2n} \cdot \frac{z^{2n}}{(2n)!}}$.

How to prove the RHS?

I see possible to come to the expansion $\pi \cot(\pi z) = \frac{1}{z}-2 \cdot \sum_{n=1}^{\infty}{\zeta(2n) z^{2n}}$ but this does not seem to be helpful enough.

Regarding straightforward approach: it is possible to find an explicit series for $z \cdot \cot(z) = z \cdot \frac{\cos(z)}{\sin(z)} = A(z)$ and then try looking for the reccurence on the coefficients of $A(z)$ but it's related with to many computational difficulties.

It is more or less reasonable way to proceed it?

Any help would be much appreciated.

hyperkahler
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    Do you know that the Bernoulli numbers appear in the Taylor series of $\frac{z}{e^z-1}$? It's not hard to get the relation from that. – Daniel Fischer Jun 07 '15 at 16:36

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By definition, $$f(z)= \frac{z}{e^z-1}=\sum_{n\geq 0}\frac{B_n}{n!}\,z^n.\tag{1}$$ By replacing $z$ with $2iz$, $$\frac{2iz}{e^{2iz}-1}=\sum_{n\geq 0}\frac{(2i)^n B_n}{n!}\,z^n.\tag{2}$$ Now: $$ z\cot z=iz\cdot\frac{e^{iz}+e^{-iz}}{e^{iz}-e^{-iz}}=iz\cdot\frac{e^{2iz}+1}{e^{2iz}-1}=iz+\frac{2iz}{e^{2iz}-1},\tag{3}$$ hence by $(2)$: $$z\cot z = iz+\sum_{n\geq 0}\frac{(2i)^n B_n}{n!}\,z^n \tag{4}$$ and since the LHS is an even function, we may drop the $z^{\text{odd}}$ parts of the RHS: $$z\cot z = \sum_{n\geq 0}\frac{(2i)^{2n} B_{2n}}{(2n)!}\,z^{2n}\tag{5}$$ as wanted.

Jack D'Aurizio
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