I would like to show the following:


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.

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

You can use Residue identity for function composition: $ \operatorname{Res}(f; h(a)) = \operatorname{Res}((f∘h)h'; a)$:

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

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.)

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

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