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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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    +1. See [here](http://en.wikipedia.org/wiki/Bernoulli_number#Asymptotic_approximation).2011-10-12
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    @Peter You misunderstand my point. There is no simple closed formula for the Bernoulli numbers, so you have to say what you want. Asymptotics? Recurrence? Double sum? You find all this in the wikipedia article on Bernoulli numbers.2011-10-12
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    @Peter What is "the usual" way to calculate Bernoulli numbers?2011-10-12