# Proof of $\lim_{n\to\infty}\frac{e^nn!}{n^n\sqrt{n}}=\sqrt{2\pi}$

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?

• – Jlamprong Jun 12 '14 at 8:12
• @Jlamprong: Yes, but what is the proof? – Riccardo.Alestra Jun 12 '14 at 8:15
• See here or here. – Ragib Zaman Jun 12 '14 at 8:18
• 1) log the expression 2) use Euler-Maclaurin formula – Alex Jun 12 '14 at 8:25
• I think it is worth noting the link between Euler's initial integral expression for the $\Gamma$ function, and the fact that $\displaystyle\int\ln x~dx=x\ln\dfrac xe$ , which, when exponentiated, becomes $\bigg(\dfrac xe\bigg)^x$ – Lucian Jun 12 '14 at 18:29

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, $$\lim_{n\to\infty}\frac{e^nn!}{n^{n+1/2}}=\sqrt{2\pi}\tag{11}$$

• I've just noticed that some of this argument can be found in this answer. – robjohn Nov 13 '14 at 20:48


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}