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how to calculate the residue of $$\frac{1}{z^{2n}} \pi \cot(\pi z)$$ at $z=0$

I know the answer is $$(2\pi i)^{2n} \frac{B_{2n}}{(2n)!}$$ but I dont know how

I saw an answer using "the coefficient extraction operator" but I dont know any thing about it

also I tried with $$ \sum_{m=0}^\infty B_{2m} (2\pi i)^{2m} \frac{z^{2m}}{(2m)!} = \pi z \cot(\pi z).$$

but I faced many problem

so what is your suggest to solve the problem ?

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mnsh
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1 Answers1

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$\cot(z)$ can be expanded as
$$ \cot (z) = \frac 1z - \frac z 3 + \dots (-1)^n \frac{B_{2n}(2z)^{2n}}{(2n)! z}\dots $$ It follows from $$\sum_{n=0}^\infty \frac{B_n z^n}{n!} = \frac{z}{e^z-1} = \frac{z}{2}\left( \coth \left ( \frac z2 \right ) - 1\right)$$

The coefficient of $\displaystyle \frac 1 z$ in $\displaystyle \frac{1}{z^{2n}} \pi \cot(\pi z)$ would be $\displaystyle (-1)^n \frac{B_{2n}(2\pi)^{2n}}{(2n)! }$ which is your residue.

S L
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