$\Gamma(n+a) \sim n^{a-1} n!$ this relation is useful to get the acceleration of a function that can be written or expressed using Gamma function. I want to know its name and the proof if possible. what I understand that it is related to Gautschi inequality, Stirling relation and strict log-convexity of Gamma function. But still can't get its name. One of the answers gives clear proof to me.
I find the answer under this question Asymptotic approximation regarding the Gamma function $\Gamma$.
We can demonstrate it with the Euler formula which gives Gamma as a limit of a product:
$\Gamma(z) = \lim_{n \to {+\infty}} \frac{n! \; n^z}{z \; (z+1)\cdots(z+n)}$
Let $n\in\mathbb{N}^*$. Considering $\Gamma(a+n+1)$, we have (by recurrence)
$\Gamma(a+n+1)=\Gamma(a).\prod_{k=0}^{n} (a+k)$
Thus
$\frac{\Gamma(a+n+1)}{n! n^a}=\Gamma(a).\frac{\prod_{k=0}^{n} (a+k)}{n! n^a}$
We recognize the inverse of the previous quotient
So $\lim_{n\rightarrow +\infty} \frac{\Gamma(a+n+1)}{n! n^a}=1$
However $\Gamma(a+n+1)=(a+n)\Gamma(a+n)$
Hence $\frac{\Gamma(a+n)}{n! n^{a-1}}=\frac{\Gamma(a+n+1)}{n! n^a}\frac{n}{n+a}$
As $\lim_{n\rightarrow +\infty} \frac{n}{n+a}=1$
$\boxed{\lim_{n\rightarrow +\infty} \frac{\Gamma(a+n)}{n! n^{a-1}}=1}$
Try to see this https://en.wikipedia.org/wiki/Stirling%27s_approximation. I think your answer is here
Consider $$y=\Gamma(n+a)\implies \log(y)=\log (\Gamma (n+a))$$ Suppose that $a\ll n$ and use expansion to get $$\log(y)=\log (\Gamma (n))+a \psi (n)+O\left(a^2\right)$$ Now, for large $n$ $$\psi (n)=\log \left({n}\right)-\frac{1}{2 n}+O\left(\frac{1}{n^3}\right)$$ $$\log(y)\sim\log (\Gamma (n))+a\log \left({n}\right)=\log \left(n^a \Gamma (n)\right)\implies y \sim n^a \Gamma (n)=n^{a-1}\Gamma(n+1)=n^{a-1}\,n!$$
For real $x$ and $0\lt s\lt1$, the inequality $$ x^{1-s}<\frac{\Gamma(x+1)}{\Gamma(x+s)}<(x+1)^{1-s}\tag1 $$ is called Gautschi's Inequality and a proof is given in this answer.
In this answer, similar methods (the log-convexity of $\Gamma$) are used to prove two extension for $0\le\alpha\le k$ $$ \left(\frac{n+\alpha-k}{n}\right)^\alpha \le\frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha} \le\left(\frac{n+k-1}{n}\right)^\alpha\tag2 $$ and $$ \left(\frac{n-k}{n}\right)^{-\alpha} \ge\frac{\Gamma(n-\alpha)}{\Gamma(n)\,n^{-\alpha}} \ge\left(\frac{n-\alpha+k-1}{n}\right)^{-\alpha}\tag3 $$ $(2)$ or $(3)$ and the Squeeze Theorem justify the limit in the question.
The similarity of $(2)$ and $(3)$ to Gautschi seems to make citing Gautschi's Inequality a good thing to do.
This answer also shows how to get the limit in this question for complex arguments (since $\Gamma(n+1)=n!$).
