Stirling's Formula and Its Simple Proof

In problem 1 (f) of tutorial note 10 we have mentioned that the power series  \[

\sum_{n=1}^\infty \frac{n!}{n^n}x^n

\] diverges when $|x|$ is the radius of convergence $e$. In fact, we can show that \[
n!\brac{\frac{e}{n}}^n\to \infty
\] by Stirling's Formula (in the simplest form), which is the following asymptotic result:

Theorem. We have $\dis n!\sim \sqrt{2\pi n} \brac{\frac{n}{e}}^n$ as $n\to \infty$.

Which actually says that \[
\limn \frac{n!}{n^{n+1/2}e^{-n}}=\sqrt{2\pi}.

\] We are going to prove this result with the help of the following simple exercises, each of them will be used exactly once.

Exercises

Problem 1. Show that for each $n\in \N$,
\[\brac{\frac{n+1}{e}}^n<n!\]

Problem 2. Show that the function \[

f(x)=\brac{1+\frac{1}{x}}^{x+\alpha}

\] is decreasing on $(0,\infty)$ when $\alpha \ge1/2$.

Problem 3. If $f:(a,b)\to \R$ is convex, then show that for every $x_1,x_2\in (a,b)$ with $x_1<x_2$, one has \[

f\brac{\frac{x_1+x_2}{2}}\leq \frac{1}{x_2-x_1}\int_{x_1}^{x_2}f(t)\,dt\leq \frac{f(x_1)+f(x_2)}{2}.

\]

Problem 4 (Walli's Formula). Define $\dis I_n=\int_0^{\pi/2}\sin^n x\,dx$, show that \[
I_{2n}=\frac{(2n-1)!!}{(2n)!!}\cdot \frac{\pi}{2}\quad\text{and}\quad I_{2n+1}=\frac{(2n)!!}{(2n+1)!!}.
\] Hence by considering $I_{2n}/I_{2n+1}$, show that \[
\limn \frac{1}{2n+1}\brac{\frac{(2n)!!}{(2n-1)!!}}^2=\frac{\pi}{2}.
\] 

 Proof of Stirling's Formula

Motivation:

By problem 1 we have\[

 \brac{1+\frac{1}{n}}^n<\frac{e^nn!}{n^n}.

\] We note that LHS increases to $e$, thus if $\dis b_n:=\frac{e^nn!}{n^n}$ also converges, we should get at least a nonzero limit. To study this, consider the quotient \[

\frac{b_{n+1}}{b_n}=e\brac{\frac{n}{n+1}}^n\iff \frac{b_{n}}{b_{n+1}}=\frac{1}{e}\brac{1+\frac{1}{n}}^n<1.

\] Therefore $\{b_n\}$ is increasing, however this limit is possibly unbounded. On the other hand, we observe that by problem 2 we can obtain a decreasing sequence by multiplying $b_n/b_{n+1}$ a factor $\dis \brac{1+\frac{1}{n}}^{1/2}=\frac{\sqrt{n+1}}{\sqrt{n}}$, namely, \[

\frac{b_n/\sqrt{n}}{b_{n+1}/\sqrt{n+1}}=\frac{1}{e}\brac{1+\frac{1}{n}}^{n+1/2}>1.

\] So instead we try to compute the limit of \[

\boxed{\dis a_n:=\frac{b_n}{\sqrt{n}}=\frac{e^nn!}{n^{n+1/2}}}

\] as at least this limit must not be unbounded.

Proof to Stirling Formula:

Now observe that \begin{align*}
1&<\frac{a_n}{a_{n+1}}=e^{-1}\brac{1+\frac{1}{n}}^{n+1/2}=\exp\bigg(-1+\Big(n+\frac{1}{2}\Big) \int_{n}^{n+1}\frac{1}{x}\,dx\bigg),
\end{align*} by problem 3 we have $\dis \int_{n}^{n+1}\frac{1}{x}\,dx\leq\frac{1}{2}\brac{\frac{1}{n}+\frac{1}{n+1}}$, therefore a simple computation yields \[

1<\frac{a_n}{a_{n+1}}<e^{\frac{1}{4}\brac{\frac{1}{n}-\frac{1}{n+1}}}.

\] Thus the first inequality tells us $\{a_n\}$ converges to a nonnegative limit $\boxed{\alpha:=\limn a_n}$. On the other hand, the second inequality tell us $\{a_ne^{-\frac{1}{4n}}\}$ increases to $\alpha$, so $\alpha >0$. By using the following form of Walli's formula (modified from problem 4) \[

\frac{(n!)^2 2^{2n}}{(2n)!}\sim \sqrt{\pi n}

\] and using $n!=a_nn^{n+1/2}e^{-n}$ we get \[

\sqrt{\pi}\leftarrow \frac{(n!)^22^{2n}}{\sqrt{n}(2n)!}=\frac{1}{\sqrt{2}} \frac{a_n^2}{a_{2n}}\to \frac{\alpha}{\sqrt{2}}.

\] Therefore we conclude that \[

\limn \frac{n!}{n^{n+1/2} e^{-n}}=\limn a_n=\alpha =\sqrt{2\pi}.\qed

\] For the ease of memorization this result is usually written as \[

\boxed{\dis n!\sim \sqrt{2\pi n} \brac{\frac{n}{e}}^n.}

\]