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I am studying for exams in complex analysis and taking a look at past papers. This comes up often or an integral of the given function along a certain curve, which is actually the same thing since the residue is all one needs.

Googling this came up but I seem to be unable to go from there.

Thanks in advance!

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Paris
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We may define Bernoulli numbers through: $$ \frac{z}{e^{z}-1}=\sum_{n\geq 0}\frac{z^n}{n!}\,B_n.\tag{1} $$ It follows that: $$ \frac{1}{e^{z}-1}=\sum_{n\geq -1}\frac{z^{n}}{(n+1)!}\,B_{n+1},\tag{2} $$ $$ \frac{-e^z}{(e^{z}-1)^2}=\sum_{n\geq -2}\frac{(n+1)\,z^{n}}{(n+2)!}\,B_{n+2},\tag{3} $$ hence the Laurent expansion of our function follows from considering the opposite of the sum between $(2)$ and $(3)$. $(3)$ follows from $(2)$ by differentiation with respect to $z$.

Jack D'Aurizio
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