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I'm looking for a proof of the following limit: $$\lim_{n\to\infty}\frac{e^nn!}{n^n\sqrt{n}}=\sqrt{2\pi}$$ This follows from Stirling's Formula, but how can it be proven?

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Existence of the Limit

The ratio of two consecutive terms is $$ \left.\frac{e^nn!}{n^{n+1/2}}\middle/\frac{e^{n-1}(n-1)!}{(n-1)^{n-1/2}}\right. =e\left(1-\frac1n\right)^{n-1/2}\tag{1} $$ Using the Taylor approximation $$ \log\left(1-\frac1n\right)=-\frac1n-\frac1{2n^2}-\frac1{3n^3}+O\left(\frac1{n^4}\right)\tag{2} $$ we can compute the logarithm of the right hand side of $(1)$: $$ 1+\left(n-\frac12\right)\log\left(1-\frac1n\right) =-\frac1{12n^2}+O\left(\frac1{n^3}\right)\tag{3} $$ Therefore, $$ \begin{align} \lim_{n\to\infty}\frac{e^nn!}{n^{n+1/2}} &=e\prod_{n=2}^\infty\left(\frac{e^nn!}{n^{n+1/2}}\middle/\frac{e^{n-1}(n-1)!}{(n-1)^{n-1/2}}\right)\\ &=e\prod_{n=2}^\infty\left(e\left(1-\frac1n\right)^{n-1/2}\right)\\ &=\exp\left(1+\sum_{n=2}^\infty\left(1+\left(n-\frac12\right)\log\left(1-\frac1n\right)\right)\right)\\ &=\exp\left(1+\sum_{n=2}^\infty\left(-\frac1{12n^2}+O\left(\frac1{n^3}\right)\right)\right)\tag{4} \end{align} $$ Since the sum on the right hand side of $(4)$ converges, so does the limit on the left hand side.


Log-Convexity of $\boldsymbol{\Gamma(x)}$

Since $\Gamma(x)$ is log-convex, $$ \begin{align} \sqrt{n}\,\Gamma(n+1/2) &\le\sqrt{n}\,\Gamma(n)^{1/2}\,\Gamma(n+1)^{1/2}\\ &=\Gamma(n+1)\tag{5} \end{align} $$ and $$ \begin{align} \Gamma(n+1) &\le\Gamma(n+1/2)^{1/2}\,\Gamma(n+3/2)^{1/2}\\ &=\sqrt{n+1/2}\,\Gamma(n+1/2)\tag{6} \end{align} $$ Therefore, $$ \sqrt{\frac{n}{n+1/2}}\le\sqrt{n}\frac{\Gamma(n+1/2)}{\Gamma(n+1)}\le1\tag{7} $$ Thus, by the Squeeze Theorem, $$ \lim_{n\to\infty}\sqrt{n}\frac{\Gamma(n+1/2)}{\Gamma(n+1)}=1\tag{8} $$


Using the Recursion for $\boldsymbol{\Gamma(x)}$

Using the identity $\Gamma(x+1)=x\Gamma(x)$, $$ \begin{align} \sqrt{n}\color{#C00000}{\frac{(2n)!}{2^nn!}}\color{#00A000}{\frac1{2^nn!}} &=\sqrt{n}\frac{\color{#C00000}{1}}{\color{#00A000}{2}}\cdot\frac{\color{#C00000}{3}}{\color{#00A000}{4}}\cdot\frac{\color{#C00000}{5}}{\color{#00A000}{6}}\cdots\frac{\color{#C00000}{2n-1}}{\color{#00A000}{2n}}\\ &=\sqrt{n}\frac{\color{#C00000}{1/2}}{\color{#00A000}{1}}\cdot\frac{\color{#C00000}{3/2}}{\color{#00A000}{2}}\cdot\frac{\color{#C00000}{5/2}}{\color{#00A000}{3}}\cdots\frac{\color{#C00000}{n-1/2}}{\color{#00A000}{n}}\\ &=\sqrt{n}\color{#C00000}{\frac{\Gamma(n+1/2)}{\Gamma(1/2)}}\color{#00A000}{\frac{\Gamma(1)}{\Gamma(n+1)}}\tag{9} \end{align} $$


Value of the Limit

Combining $(8)$ and $(9)$ yields $$ \begin{align} \lim_{n\to\infty}\left(\frac{e^nn!}{n^{n+1/2}}\right)^{-1} &=\lim_{n\to\infty}\frac{e^{2n}(2n)!}{(2n)^{2n+1/2}}\left(\frac{n^{n+1/2}}{e^nn!}\right)^2\\ &=\lim_{n\to\infty}\frac{\sqrt{n}}{\sqrt2\,4^n}\frac{(2n)!}{(n!)^2}\\ &=\lim_{n\to\infty}\frac1{\sqrt2}\frac{\Gamma(1)}{\Gamma(1/2)}\sqrt{n}\frac{\Gamma(n+1/2)}{\Gamma(n+1)}\\ &=\frac1{\sqrt{2\pi}}\tag{10} \end{align} $$ Therefore, $$ \lim_{n\to\infty}\frac{e^nn!}{n^{n+1/2}}=\sqrt{2\pi}\tag{11} $$

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  • $\begingroup$ I've just noticed that some of this argument can be found in this answer. $\endgroup$ – robjohn Nov 13 '14 at 20:48
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$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,{\rm e}^{#1}\,} \newcommand{\fermi}{\,{\rm f}} \newcommand{\floor}[1]{\,\left\lfloor #1 \right\rfloor\,} \newcommand{\half}{{1 \over 2}} \newcommand{\ic}{{\rm i}} \newcommand{\iff}{\Longleftrightarrow} \newcommand{\imp}{\Longrightarrow} \newcommand{\pars}[1]{\left(\, #1 \,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\pp}{{\cal P}} \newcommand{\root}[2][]{\,\sqrt[#1]{\vphantom{\large A}\,#2\,}\,} \newcommand{\sech}{\,{\rm sech}} \newcommand{\sgn}{\,{\rm sgn}} \newcommand{\totald}[3][]{\frac{{\rm d}^{#1} #2}{{\rm d} #3^{#1}}} \newcommand{\ul}[1]{\underline{#1}} \newcommand{\verts}[1]{\left\vert\, #1 \,\right\vert}$

With the Stirling's Approximation for $n!$ $\ds{\approx \root{2\pi}n^{n + 1/2}\expo{-n}}$

\begin{align} \color{#66f}{\large\lim_{n\ \to\ \infty}{\expo{n}n! \over n^{n}\root{n}}} =\lim_{n\ \to\ \infty} {\expo{n}\color{#c00000}{\root{2\pi}n^{n + 1/2}\expo{-n}} \over n^{n + 1/2}} =\color{#66f}{\large\root{2\pi}} \approx {\tt 2.5066} \end{align}

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