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On the wikipedia page for Gamma function I saw an interesting formula $$ \lim_{n\to \infty} \frac{\Gamma(n+\alpha)}{\Gamma(n)n^\alpha} = 1 $$ for all $\alpha\in\Bbb C$. I couldn't find the source of this and searching here in MSE didn't provide the result I want.

Could anyone show me how this formula is derived?

I'm very inexperienced with properties/identities of $\Gamma$ so forgive me if this question is trivial.

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  • $\begingroup$ Use Stirlings Approximation. It's probably on that Wikipedia page. $\endgroup$ – Brevan Ellefsen Jan 16 at 22:46
  • $\begingroup$ So Stirlings approximation also works in complex case? Thank you, I didn't know that before. $\endgroup$ – BigbearZzz Jan 16 at 22:47
  • $\begingroup$ This follows from Gautschi's Inequality, which can be proven using the log-convexity of the Gamma function. $\endgroup$ – robjohn Jan 17 at 4:40
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The most usual derivation of this would involve the Stirling-Laplace asymptotic for $\Gamma(s)$. I'm mildly surprised that this wasn't explicitly worked out in Wiki, or some other easily accessible places on-line.

In fact, a much simpler approach obtains (a stronger version of) this asymptotic via "Watson's Lemma", which is itself easy to completely prove from simple things, going back over 100 years. In various places in the literature, the lemma is in fact called "the oft-reproven Watson's lemma". :)

The case you mention is a simple corollary of the very first example I wrote out in some notes on asymptotic expansions: http://www.math.umn.edu/~garrett/m/mfms/notes_2013-14/02d_asymptotics_of_integrals.pdf

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    $\begingroup$ Thank you for you answer. By the way, your link doesn't work for me for some reason. $\endgroup$ – BigbearZzz Jan 16 at 22:53
  • $\begingroup$ Oop, let me try to repair the link... $\endgroup$ – paul garrett Jan 16 at 23:03
  • $\begingroup$ I think it works now... :) $\endgroup$ – paul garrett Jan 16 at 23:04
  • $\begingroup$ Now it also works perfectly for me. Thanks! $\endgroup$ – BigbearZzz Jan 16 at 23:16
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$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$

Stirling Asymptotic: $\ds{N! \sim \root{2\pi}\, N^{N + 1/2}\expo{-N}}$ as $\ds{\verts{N} \to \infty}$.

\begin{align} \left.\lim_{n \to \infty}{\Gamma\pars{n + \alpha} \over \Gamma\pars{n}n^{\alpha}}\,\right\vert_{\ \alpha\ \in\ \mathbb{C}} & = \lim_{n \to \infty}{\pars{n + \alpha - 1}! \over \pars{n - 1}!\, n^{\alpha}} \\[5mm] & = \lim_{n \to \infty}{\root{2\pi}\pars{n + \alpha - 1}^{n + \alpha - 1/2}\expo{-\pars{n + \alpha - 1}} \over \bracks{\root{2\pi}\pars{n - 1}^{n - 1/2}\expo{-\pars{n - 1}}}\, n^{\alpha}} \\[5mm] & = \lim_{n \to \infty}{n^{n + \alpha - 1/2}\, \bracks{1 + \pars{\alpha - 1}/n}^{n + \alpha - 1/2}\,\expo{-\alpha} \over \bracks{n^{n - 1/2}\pars{1 - 1/n}^{n - 1/2}}\, n^{\alpha}} \\[5mm] & = \expo{-\alpha}\lim_{n \to \infty} {\bracks{1 + \pars{\alpha - 1}/n}^{n} \over \pars{1 - 1/n}^{n}} \\[5mm] & = \expo{-\alpha}\,{\expo{\alpha - 1} \over \expo{-1}} = \bbx{1} \end{align}

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Real $\boldsymbol{\alpha}$

The log-convexity of the Gamma function is shown in this answer.

Suppose that $0\le\alpha\le k\in\mathbb{Z}$, then using the recurrence relation for $\Gamma$, $$ \begin{align} \Gamma(n+\alpha) &\le\Gamma(n)^{1-\alpha/k}\,\Gamma(n+k)^{\alpha/k}\\ &\le\Gamma(n)^{1-\alpha/k}\left(\Gamma(n)\,(n+k)^k\right)^{\alpha/k}\\ &=\Gamma(n)\,(n+k)^\alpha\tag1 \end{align} $$ and $$ \begin{align} \Gamma(n) &\le\Gamma(n+\alpha-k)^{\alpha/k}\Gamma(n+\alpha)^{1-\alpha/k}\\[6pt] &\le\left(\frac{\Gamma(n+\alpha)}{(n+\alpha-k)^k}\right)^{\alpha/k}\Gamma(n+\alpha)^{1-\alpha/k}\\ &=\frac{\Gamma(n+\alpha)}{(n+\alpha-k)^\alpha}\tag2 \end{align} $$ Then we have $$ \left(\frac{n+\alpha-k}{n}\right)^\alpha \le\frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha} \le\left(\frac{n+k}{n}\right)^\alpha\tag3 $$ and by the Squeeze Theorem, for $\alpha\ge0$, $$ \lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha}=1\tag4 $$ Substituting $n\mapsto n-\alpha$ in $(3)$, multiplying by $\left(\frac{n-\alpha}n\right)^\alpha$, and taking reciprocals, we get $$ \left(\frac{n-k}{n}\right)^{-\alpha} \ge\frac{\Gamma(n-\alpha)}{\Gamma(n)\,n^{-\alpha}} \ge\left(\frac{n-\alpha+k}{n}\right)^{-\alpha}\tag5 $$ and by the Squeeze Theorem, for $\alpha\ge0$, $$ \lim_{n\to\infty}\frac{\Gamma(n-\alpha)}{\Gamma(n)\,n^{-\alpha}}=1\tag6 $$


Complex $\boldsymbol{\alpha}$

Unfortunately, I have not found a way to make the log-convexity argument that works for $\alpha\in\mathbb{R}$ work for $\alpha\in \mathbb{C}$. About the best I can see, is to use Stirling's Approximation. $$ \Gamma(n)\sim\sqrt{\frac{2\pi}n}\frac{n^n}{e^n}\tag7 $$ Applying $(7)$ to $\Gamma(n+\alpha)$ and $\Gamma(n)$, we get $$ \frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha}\sim\sqrt{\frac{n}{n+\alpha}}\frac{\left(1+\frac\alpha{n}\right)^{n+\alpha}}{e^\alpha}\tag8 $$ which yields $$ \lim_{n\to\infty}\frac{\Gamma(n+\alpha)}{\Gamma(n)\,n^\alpha}=1\tag9 $$

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  • $\begingroup$ Fyi, the Bohr-Mollerup Wikipedia page proves the real case as a direct corollary. $\endgroup$ – adfriedman Jan 18 at 2:37
  • $\begingroup$ @adfriedman: that is not surprising since both are using the log-convexity of $\Gamma$. $\endgroup$ – robjohn Jan 18 at 4:15
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As said, Stirling approximation is the key.

Considering $$y=\frac{\Gamma(n+a)}{\Gamma(n)\,n^a}\implies \log(y)=\log (\Gamma (a+n))-\log (\Gamma (n))-a \log(n)$$ use Stirling approximation $$\log (\Gamma (p))=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )-\log \left({p}\right)\right)+\frac{1}{12 p}-\frac{1}{360 p^3}+O\left(\frac{1}{p^5}\right)$$ Just apply it and continue with Taylor series to get $$\log(y)=\frac{(a-1) a}{2 n}-\frac{(a-1) a (2 a-1)}{12 n^2}+\frac{(a-1)^2 a^2}{12 n^3}+O\left(\frac{1}{n^4}\right)$$ Continue with Taylor $$y=e^{\log(y)}=1+\frac{(a-1) a}{2 n}+\frac{(a-2) (a-1) a (3 a-1)}{24 n^2}+O\left(\frac{1}{n^3}\right)$$

Using it for $n=100$ and $a=5+7i$ the "exact value" would be $(0.798463+ 0.143902 \,i)$ while the above approximation would give $\frac{31947}{40000}+\frac{1057 }{7500}i\approx (0.798675 +0.140933 i)$.

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