About Beta function $B(\alpha,r\alpha +1)‎\rightarrow‎ 0$

Question

I want to show that $$B(\alpha,r\alpha +1)‎\rightarrow‎ 0$$ when $r‎\rightarrow‎ \infty$ and $0< \alpha <1$.

with thanks

Answer 1

From $$ \mathrm{B}(x,y)=\frac{\Gamma (x)\Gamma (y)}{\Gamma (x+y)} $$ you get $$ \mathrm{B}(\alpha,r\alpha +1)=\frac{\Gamma (\alpha)\Gamma (r\alpha+1)}{\Gamma ((r+1)\alpha+1)} $$ then use Stirling's formula, as $z \to \infty$, $$ \Gamma(z) = \sqrt{2\pi} z^{z - 1/2} e^{-z} (1 + O(1/z)) $$ to get, as $\alpha \to \infty$,

$$ \mathrm{B}(\alpha,r\alpha +1)\sim\frac{\Gamma (\alpha)}{\alpha^\alpha}\frac1{r^\alpha},\quad \quad 0<\alpha<1. $$

The announced result follows.