What is $\lim\limits_{n\to\infty}\frac{\Gamma(n+1)\Gamma(\alpha-n+1)}{n^{\alpha+1} }$

Question

Let $\alpha\in\mathbb{R}$ then, I would like to compute $$\lim_{n\to\infty}\frac{\Gamma(n+1)\Gamma(\alpha-n+1)}{n^{\alpha+1} }$$ where $\Gamma$ is the standard gamma function.

Answer 1

By Gamma extension formula we have

$$ \Gamma(\varepsilon-n) =(-1)^n\frac{ \Gamma(1-\varepsilon)\Gamma(\varepsilon)}{ \Gamma(n+1-\varepsilon)}~~~~0<\varepsilon<1~~n\in \Bbb N$$ Also if $\alpha$ is an integer then the limit won't makes any sense, since the Gamma function does exists for negative integers. Therefore looking at the floor of $\alpha $, we have, $$\alpha =m +\varepsilon~~~0<\varepsilon<1, ~~m\in \Bbb Z.$$

Now for $n>m,$ write

$$\color{red}{ \Gamma(\alpha-n+1) = \Gamma(\varepsilon-(n-m-1)) = (-1)^{n-m}\frac{ \Gamma(1-\varepsilon)\Gamma(\varepsilon)}{ \Gamma(n-m-\varepsilon)}}$$

That is, $$\color{red}{ \Gamma(\alpha-n+1) = (-1)^{n-m}\frac{ \Gamma(1-\varepsilon)\Gamma(\varepsilon)}{ \Gamma(n-\alpha)}}$$

Hence for $n>m,$

$$\color{blue}{ \Gamma(n+1)\Gamma(\alpha-n+1)n^{-\alpha-1} = (-1)^{n-m}\Gamma(1-\varepsilon)\Gamma(\varepsilon)\frac{\Gamma(n)n^{-\alpha}}{ \Gamma(n-\alpha)}.}$$

Nevertheless, from this On a reference for $ \lim _{n\to \infty }{\frac {\Gamma (n+\alpha )}{\Gamma (n)n^{\alpha }}}=1,\qquad \alpha \in \mathbb {C} $ We have that,

$$\lim_{n\to \infty}\frac{\Gamma(n)n^{-\alpha}}{ \Gamma(n-\alpha)} = 1.$$

Conclusion, $$\color{blue}{ \lim_{n\to \infty}\Gamma(n+1)\Gamma(\alpha-n+1)n^{-\alpha-1} ~~~~~\text{Does not exist for any $\alpha\in \Bbb R$}}$$