# Computing residue involving Bernoulli numbers

I would like to show the following:

$$\operatorname{Res}_{z=0}\left(\frac{e^{nz}}{(1-e^{-z})^{m+1}}\right)=\binom{n+m}m$$

Basically I have to show that the $$z^{-1}$$ term in $$\left(\sum_{k=0}^\infty \frac{(nz)^k}{k!}\right)\left(\frac1z +\frac12 +\sum_{i=1}^\infty \frac{(-1)^{i-1}B_i}{(2i)!}x^{2i}\right)^{m+1}$$

is given by $$\binom{n+m}m$$. I am not sure about how to simplify the Bernoulli numbers in the products and sums.

• I should give some context here. This is basically a calculation in algebraic geometry about the Hirzebruch-Riemann-Roch formula applied on $\mathcal O_{\mathbb P^m}(n)$. – lEm Feb 4 at 13:00

$$\operatorname{Res}_{z=a}\left(f(h(z))h'(z)\right)=\operatorname{Res}_{w=h(a)}(f(w))\;.$$

In your case we can take $$h(z)=\mathrm e^z$$ and

$$f(w)=\frac{w^{n-1}}{\left(1-w^{-1}\right)^{m+1}}=\frac{((w-1)+1)^{n+m}}{(w-1)^{m+1}}\;.$$

Then the residue of $$f\circ h$$ at $$a=0$$ is the residue of $$f$$ at $$h(a)=1$$, which is the coefficient of $$x^m$$ in $$(x+1)^{n+m}$$, which is indeed $$\binom{n+m}m$$.

(A similar result is shown at Residue of composite functions, but only for simple poles.)

• Thank you very much. It is simpler than I thought! – lEm Feb 4 at 13:41
• @lEm: If a result is that simple, it usually has a correspondingly simple derivation :-) – joriki Feb 4 at 13:54
• That's a very useful identity for calculating residues. I'm surprised I've never seen it before. – Random Variable Feb 4 at 20:02
• @RandomVariable: Same here :-) – joriki Feb 4 at 20:02