Its relation to $\Gamma$

We note that

$$\gamma = \lim_{n} \left( \sum_{k=1}^{n}\frac{1}{k} - \log n \right)$$

exists (the sequence in the brackets is decreasing and positive). It is called Euler's constant. We now put

$$\Delta (z) = z \exp (\gamma z) \prod_{n=1}^{\infty}(1+z/n)\exp (-z/n),$$

and we claim that

$$\frac{1}{\Gamma (z)}=\Delta (z), \qquad \mbox{for } z \in \mathbb C \setminus \{ 0,-1,-2, \cdots \} .$$

To see this we define

$$G_n(z)=\frac{n!n^z}{z(z+1)\cdots(z+n)}$$

and notice that

$$G_n(z)z \exp (\gamma z) \prod_{k=1}^{n}(1+z/k)\exp (-z/k) = \... ...amma -\sum_{k=1}^n\frac{11}{k} +\log n\right) z\right) \to 1$$

uniformly on compact subsets of $\mathbb C \setminus \{ 0,-1,-2, \cdots \}$. Thus

$$G_n(z)\to G(z)\equiv \frac{1}{\Delta (z)}.$$

Furthermore,

$$G_n(z+1)=\frac{n!n^zn}{(z+1)\cdots(z+n+1)}= G_n(z)\frac{nz}{z+n+1}$$

and

$$G_n(1)=\frac{n}{n+1}.$$

Letting n tend to infinity we obtain

$$G(z+1)=zG(z) \qquad G(1)=1.$$

Finally, $\vert G(z)\vert\leq G(\Re z)$, so Wielandt's theorem can be applied and we get

$$G(z)=\Gamma (z).$$