$\Gamma(s)= \lim_{n \to \infty} \frac{n^s n!}{s(s+1)...(s+n)}$ , the product formula of Gamma function .

Question

Prove that $$\Gamma(s)= \lim_{n \to \infty} \frac{n^s n!}{s(s+1)...(s+n)}$$ Whenever $s \neq 0,-1,-2,...$

My attempt :
By applying product formula for $\frac{1}{\Gamma}$ , $$\Gamma(s)=\lim_{n\to \infty} \frac{n!}{s(s+1)...(s+n)}e^{s(1+\frac12+...+\frac1n-\gamma)}$$ Notice that $1+\frac12+...+\frac1n-\gamma\to \log n$ , so it seems that we then get the desired conclusion . But whenever $Re(s)\gt0$ , $n^s \to \infty$ , we can not have $$\Gamma(s)=\lim_{n\to \infty} \frac{n!}{s(s+1)...(s+n)}\lim_{n\to \infty} e^{s(1+\frac12+...+\frac1n-\gamma)}$$
Let $A(s)=\lim_{n\to \infty} \frac{n!}{s(s+1)...(s+n)}n^s$ , I want to prove this exercise by showing the following statement.
1) $A(s)$ defines a meromorphic function with simple poles at $0,-1,-2,...$ and nowhere else.
2)$A(s)=\Gamma (s)$ whenever $s \in (\frac14,\frac13)$ .

To show 2) , we need some estimate of $\gamma$ , $$\sum_{n=1}^N \frac1n-\log N =\sum_{n=1}^{N-1}\int_n^{n+1} \frac1n-\frac1x \,dx +\frac1N$$ Then we may write $a_n=\int_n^{n+1} \frac1n-\frac1x \,dx $ and $\gamma=\sum_1^{\infty}a_n$ . Moreover , $a_n \le \frac{1}{n(n+1)}$
Notice that whenever $s\in (\frac14,\frac13)$ , $\frac{n!}{s(s+1)...(s+n)}$ is bounded by some fixed $M$ . $$\Gamma(s)-A(s)\le M\lim_{n\to \infty} |e^{s(1+\frac12+...+\frac1n-\gamma)}-e^{s \log n}|$$ apple mean-value theorem we have $$|e^{s(1+\frac12+...+\frac1n-\gamma)}-e^{s \log n}|\le s|1+\frac12+...+\frac1n-\gamma-\log n||e^{s*2 \log n}|$$ $$\le sn^{\frac23} |\sum_{k=n}^{\infty} a_k+\frac1n|\le sn^{\frac23} |\sum_{k=n}^{\infty} \frac{1}{k(k+1)}+\frac1n|\le \frac{2sn^{\frac23}}{n} \to 0$$ And we complete the proof of 2) . To prove 1) , we need to prove that for every compact subset $\Omega$ of $C-\{0,-1,-2,...\}$ , $f_n=\frac{n!}{s(s+1)...(s+n)}n^s$ converges uniformly . I have no idea how to deal with this .

Answer 1

A much more straightforward proof from the given definition for $\Re(s)>0$:

$$f_n(t):=\begin{cases}0,&t>n\\\left(1-\frac tn\right)^n,&0\le t\le n\end{cases}$$

\begin{align}\Gamma(s):\!&=\int_0^\infty t^{s-1}e^{-t}~\mathrm dt\\&=\lim_{n\to\infty}\int_0^\infty t^{s-1}f_n(t)~\mathrm dt\tag{DCT}\\&=\lim_{n\to\infty}\int_0^nt^{s-1}f_n(t)~\mathrm dt\\&=\lim_{n\to\infty}\int_0^nt^{s-1}\left(1-\frac tn\right)^n~\mathrm dt\\&=\lim_{n\to\infty}n^s\int_0^1u^{s-1}\left(1-u\right)^n~\mathrm du\tag{$t=nu$}\\&=\lim_{n\to\infty}n^s\prod_{k=1}^n\frac k{k+s-1}\tag{Induction over $n$}\end{align}

The last expression being your limit in product form. One can then verify the last form to satisfy the recursive equation $\Gamma(s+1)=s\Gamma(s)$ and hence extends to $\Re(s)\le0$ as the Gamma function.

Answer 2

$\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \begin{align} &\bbox[10px,#ffd]{\lim_{n \to \infty}{n^{s}\, n! \over s\pars{s + 1}\ldots\pars{s + n}}} = \lim_{n \to \infty}{n^{s}\, n! \over s^{\overline{n + 1}}} = \lim_{n \to \infty}{n^{s}\, n! \over \Gamma\pars{s + n + 1}/\Gamma\pars{s}} \\[5mm] = &\ \Gamma\pars{s}\lim_{n \to \infty}{n^{s}\, n! \over \pars{s + n}!} \\[5mm] = &\ \Gamma\pars{s}\lim_{n \to \infty}{n^{s}\bracks{\root{2\pi}n^{n + 1/2}\expo{-n}} \over \root{2\pi}\pars{n + s}^{n + s + 1/2} \expo{-\pars{n + s}}}\qquad\pars{~Stirling\ Asymptotic~} \\[5mm] = &\ \Gamma\pars{s}\lim_{n \to \infty}{n^{s + n + 1/2}\,\expo{s} \over n^{n + s 1/2}\,\pars{1 + s/n}^{n + s + 1/2}} = \Gamma\pars{s}\lim_{n \to \infty}{\expo{s} \over \pars{1 + s/n}^{n}} \\[5mm] = &\ \Gamma\pars{s}\,{\expo{s} \over \expo{s}} = \bbx{\Gamma\pars{s}} \end{align}

Answer 3

To prove uniform convergence of $f_n$ on a compact set $\Omega \in \mathbb{C} \setminus \{0,-1,-2,\ldots\}$ note that

$$\tag{*}f_n(s) = \frac{n^s n!}{s(s+1)...(s+n)} = \frac{n^s}{s} \prod_{k=1}^n \left(1 + \frac{s}{k} \right)^{-1} \\ = \left(\frac{n}{n+1} \right)^s \frac{1}{s}\prod_{k=1}^n \left(1 + \frac{1}{k} \right)^{s}\left(1 + \frac{s}{k} \right)^{-1} $$

where the last equality follows from

$$\prod_{k=1}^n \left(1 + \frac{1}{k} \right)^{s} = \left[ \prod_{k=1}^n \left(\frac{k+1}{k}\right) \right]^s = \left(\frac{(n+1)!}{n!}\right)^s = (n+1)^s$$

Since $\frac{1}{2} \leqslant \frac{n}{n+1} < 1$ it can be shown that $\left(\frac{n}{n+1} \right)^s \frac{1}{s}$ is uniformly bounded on $\Omega$. Thus, uniform convergence of $f_n$ holds if the product on the RHS of (*) is uniformly convergent.

We can apply a general theorem for complex infinite products that states that the product $\prod [1 + g_n(s)]$ is uniformly and absolutely convergent on a compact set if $\sum |g_n(s)|$ is uniformly convergent. This holds here because for large $K$ we have

$$\left(1 + \frac{1}{k} \right)^{s}\left(1 + \frac{s}{k} \right)^{-1} = 1 + \frac{s((s-1)}{2k^2} + \mathcal{O}(k^{-3})$$

Thus, the product on the RHS of (*) and, consequently $f_n$ are uniformly convergent on the compact set $\Omega$ that avoids singularities of the gamma function.