Calculate $‎\lim‎_{ ‎r\rightarrow ‎\infty‎}‎‎\frac{\Gamma(r\alpha)}{\Gamma((r+1)\alpha)}‎‎$

Question

I need to calculate limit

$$‎\lim‎_{ ‎r\rightarrow ‎\infty‎}‎‎\frac{\Gamma(r\alpha)}{\Gamma((r+1)\alpha)}‎‎$$

where $0<\alpha <1$ and $\Gamma(.)$ is Gamma function.

with thanks in advance.

Answer 1

For $0<\alpha<1$ and $z\to\infty$ we have $$\frac{\Gamma(z+a)}{\Gamma(z)}= z^{a}\left(1+O\left(\frac1z\right)\right).\tag{1}$$ Therefore $$\frac{\Gamma(r\alpha)}{\Gamma(r\alpha+\alpha)}\sim \left(r\alpha\right)^{-\alpha}$$ and the limit is $0$.

P.S. The asymptotics (1) can be derived from Stirling's approximation for the gamma function. However there is an easy heuristic way to derive it: if $a\in\mathbb N$, then $$\frac{\Gamma(z+a)}{\Gamma(z)}=z(z+1)\ldots(z+a-1)\sim z^a.$$

Answer 2

If we assume $r>\frac{1}{\alpha}$ we have: $$\frac{\Gamma(\alpha)\Gamma(r\alpha)}{\Gamma((r+1)\alpha)}=\int_{0}^{1}z^{\alpha-1}(1-z)^{r\alpha-1}\,dz\leq \int_{0}^{+\infty}z^{\alpha-1} e^{(1-r\alpha)z}\,dz = \frac{\Gamma(a)}{(r\alpha-1)^\alpha}$$ hence the limit is zero.