Laurent expansion of the gamma function at negative integers

Question

It is known that the gamma function has the non-positive integers as its simple poles. And we know the residues at these poles.

$$ \operatorname*{Res}_{z=-n}\Gamma(z) = \frac{(-1)^n}{n!} . $$

This means, its Laurent expansion around $z = -n$ is of the form

$$ \Gamma(z) = \frac{(-1)^n}{n!} \frac{1}{z+n} + C_0 + C_1 (z+n) + \cdots . $$

The question is, what is the value of $C_0$, or even $C_1$?

Answer 1

The expansion of $\Gamma(z-n)$ around $z=0$ can be obtained from known $$\Gamma(1+z)=1-\gamma z+\gamma_2 z^2+\ldots,\quad\gamma_2=\frac{1}{2}\Big(\gamma^2+\frac{\pi^2}{6}\Big)$$ using properties of $\Gamma$ and partial fractions: $$\frac{\Gamma(z-n)}{\Gamma(1+z)}=\prod_{k=0}^{n}\frac{1}{z-k}=\frac{(-1)^n}{n!}\sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^k}{z-k};$$ expanding each term using geometric series, we get $$\frac{\Gamma(z-n)}{\Gamma(1+z)}=\frac{(-1)^n}{n!}\left(\frac{1}{z}+\sum_{r=0}^{\infty}A_r z^r\right),\quad A_r=\sum_{k=1}^{n}\binom{n}{k}\frac{(-1)^{k-1}}{k^{r+1}}$$ (actually $A_0=\sum_{k=1}^{n}\frac{1}{k}$ is just $n$-th harmonic number; here are further details).

Thus $C_0=\dfrac{(-1)^n}{n!}(A_0-\gamma)$ and $C_1=\dfrac{(-1)^n}{n!}(A_1-\gamma A_0+\gamma_2)$.