3

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.

Guy Fsone
  • 23,903
user301395
  • 541
  • have you tried the stirling approximation? – Jürg W. Spaak Oct 20 '17 at 11:55
  • @JürgMerlinSpaak, How would you apply Stirling on the negative real axis? – Antonio Vargas Oct 20 '17 at 12:03
  • In order to avoid evaluating the $\Gamma$ function at a pole, you have to assume $\alpha\not\in\mathbb{N}$, too. – Jack D'Aurizio Oct 20 '17 at 12:08
  • @AntonioVargas agreed, doesn't really help... – Jürg W. Spaak Oct 20 '17 at 12:12
  • 1
    Using Euler reflection formula and Stirling approximation, you can check that $$ \frac{\Gamma(n+1)\Gamma(\alpha-n+1)}{n^{\alpha+1}} = (-1)^{n-1} \frac{\pi}{\sin \pi \alpha} + \mathcal{O}(n^{-1}). $$ So the limit does not exist for any $\alpha \in \mathbb{R}$. – Sangchul Lee Oct 24 '17 at 05:42
  • @SangchulLee the answer below says the does exist for none integer and the limit is 0 check the result and tell me – Guy Fsone Oct 30 '17 at 10:27
  • @GuyFsone, At the beginning of your answer, the formula should be corrected as $$ \Gamma(\varepsilon-n) = (-1)^n\frac{ \Gamma(1-\varepsilon)\Gamma(\varepsilon)}{ \Gamma(n+1-\varepsilon)}. $$ Notice the denominator for the difference. This matches with the Euler reflection formula as well as numerical computations. Although I haven't checked the rest, I guess that this fix will lead to the same answer. – Sangchul Lee Oct 30 '17 at 10:35
  • Yes thank your right for the formula. But the result remain the same. nothing change. – Guy Fsone Oct 30 '17 at 10:39
  • @GuyFsone, Starting from the formula above, we get $$ \frac{\Gamma(n+1)\Gamma(\alpha-n+1)}{n^{\alpha+1}} = (-1)^n \Gamma(\alpha+1)\Gamma(-\alpha) \underbrace{ \frac{\Gamma(n+1)}{\Gamma(n-\alpha)n^{1+\alpha}} }_{=:b_n}. $$ Then you can show that $b_n \to 1$ as $n\to\infty$. (The link in your answer does prove this.) So something does really change. – Sangchul Lee Oct 30 '17 at 10:39
  • 1
    Thank very much I have updated my answer. That sign was extremely relevant. And rather your answer is correct as well – Guy Fsone Oct 30 '17 at 10:55

1 Answers1

4

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

Guy Fsone
  • 23,903