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Over here we had a similar looking question:

How do I evaluate the Complex Integral $(z^n)/(e^z - 1)$ using residue theory?

and the answer was that (rather unremarkably) we could remove the singularity made by $1/(e^z-1)$ for $z=0$ and conclude Res$((z^n)/(e^z-1))=0$ for all $n>0$ and Res$(1/(e^z-1))=1$.

I think this residue calculation becomes rather more interesting when we put a $-$ in front of the $n$. Then we have an (n+1)th order pole at $z=0$. Can we calculate the residue other than with this limit formula?

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    Whats wrong with this limit formula? – 2011-10-12

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The coefficients of the power series $z/(e^z-1)$ are by definition the Bernoulli numbers (divided by a factorial), so they are the residues of your function.

Now the question remains what you mean by "calculate", i.e. what kind of answer do you seek?

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    @Peter What is "the usual" way to calculate Bernoulli numbers? – 2011-10-12