# Ratio of close gamma functions

Is there any way I can calculate this ratio more quickly than evaluating the (potentially very large) numerator and denominator?

$$\frac{\Gamma(z+\alpha)}{\Gamma(z)}$$

I'm specifically interested in the case where $$0 < \alpha < 1$$

I derived a formula for the special case where $$\alpha = \frac{1}{2}$$:

$$\Gamma(n + \frac{1}{2}) = \binom{n - \frac{1}{2}}{n}n!\sqrt{\pi}$$ $$\frac{\Gamma(n + \frac{1}{2})}{\Gamma(n + 1)} = \binom{n - \frac{1}{2}}{n}\sqrt{\pi}$$ $$\frac{\Gamma(n + 1)}{\Gamma(n + \frac{1}{2})} = \frac{1}{\binom{n - \frac{1}{2}}{n}\sqrt{\pi}}$$ $$\frac{\Gamma(z + \frac{1}{2})}{\Gamma(z)} = \frac{1}{\binom{z - 1}{z - \frac{1}{2}}\sqrt{\pi}}$$

and the obvious case where $$\alpha = 1$$:

$$\frac{\Gamma(z+1)}{\Gamma(z)} = z$$

Is there a more general formula that encompasses any $$\alpha$$?

• a way to go is defining $G(z,\alpha ):=\frac{\Gamma (z)}{\Gamma (z+\alpha )}$, then you want to estimate such function $G$ for $|\alpha |<1$ and some fixed $z$. For this task I will try to search about some of the many representations of the $\Gamma$ function that simplifies/makes efficient enough this task – Masacroso Oct 29 '19 at 16:08

There is no general form, since that would imply $$\Gamma$$ had a "closed form" for any values. It is possible to approximate the ratio as:
$$\frac{\Gamma(z+\alpha)}{\Gamma(z)}\underset{z\to\infty}\sim z^\alpha$$
which holds regardless of $$\alpha$$ and may be a direct consequence of the definition of $$\Gamma$$ depending on how you choose to define it.