I'm currently dealing with a problem my professor raised (since I just studied the Mittag-Leffler's Partial Fractions Theorem). The problem is to derive a partial fractions decomposition of the Gamma Function that displays its (simple) poles and principal parts.
So in my textbook, the Gamma Function is defined as the following limit:
$$\lim_{n \to \infty}\frac{n!\ n^z}{z(z+1)...(z+n)}$$
where $z$ is an arbitrary complex number that is not $0, -1, -2,...$
With the residue of each pole $z_v=-v,\ v=0,1,2,...$ as
$$a_{-1}=\frac{(-1)^v}{v!}$$
We have the principal parts of the Gamma Function at each pole $z_v$ as
$$h_v(z)=\frac{(-1)^v}{v!\ (z+v)}$$
And so it is sufficient to show that
$$\sum_{v=1}^{\infty} \left[h_v(z)-g_v(z) \right]$$ is uniformly convergent on $|z|\leq R,\ R>0$ , where $g_v(z)$ is defined to be the first few finite terms of the series expansion of $h_v(z)$ around $z=0$.
By considering $v$ large enough so that $|z_v|=v>2R$ and taking $g_v(z)=0$, we have that for $|z|\leq R$
$$\left|\frac{(-1)^v}{v!\ (z+v)} \right|=\frac{1}{v!\ |z+v|}<\frac{2}{v!\ v}$$
So the series above converges uniformly and hence there exists an entire function $G(z)$ such that
$$\Gamma(z) = G(z)+\frac{1}{z} + \sum_{v=1}^{\infty}\frac{(-1)^v}{v!\ (z+v)}$$
And now I am stuck at finding $G(z)$. Anyone can provide some help?