Proof of Stirling’s Formula

Stirling’s formula is very useful in all kinds of asymptotic analysis. Here we present one of many proofs.

First, Stirling’s formula says that

$$\begin{equation} \lim_{n\to\infty}\frac{n!}{\sqrt{2\pi n}(n/e)^n} = 1\,. \end{equation}$$

We start from the fact that $n! = \int_0^\infty x^ne^{-x}\operatorname{d}x$. Then do the substitution $x = t\sqrt{n}+n$. We get the representation

$$\begin{equation} n! = \frac{n^n\sqrt{n}}{e^n}\int_{-\sqrt{n}}^\infty \left(1+\frac{t}{\sqrt{n}}\right)^n e^{-t\sqrt{n}}\operatorname{d}t\,. \end{equation}$$

We denote $f_n(t) = \chi_{[-\sqrt{n},\infty)}(t)(1+t/\sqrt{n})^n e^{-t\sqrt{n}}$. Then we only need to show that the integral $\int_{-\infty}^\infty f_n(t)\operatorname{d}t$ equals $\sqrt{2\pi}$. This comes from the fact that

$$\begin{equation} \int_{-\infty}^\infty e^{-t^2/2}\operatorname{d}t = \sqrt{2\pi}\,. \end{equation}$$

So we want to show that

$$\begin{equation} \lim_{n\to\infty} \chi_{[-\sqrt{n},\infty)}(t)\left(1+\frac{t}{\sqrt{n}}\right)^n e^{-t\sqrt{n}} = e^{-t^2/2}\,. \end{equation}$$

Assuming that $\vert t\vert < \sqrt{n}$, we compare the natural log of two sides. By the Taylor series of $\ln(1+x)$, we have

$$\begin{align} \ln\left(\left(1+\frac{t}{\sqrt{n}}\right)^n e^{-t\sqrt{n}}\right) & = n\ln\left(1+\frac{t}{\sqrt{n}}\right)-t\sqrt{n} \\ & = n\left(\frac{t}{\sqrt{n}}-\frac{t^2}{2n}+o(1/n)\right) - t\sqrt{n} \\ & = -\frac{t^2}{2} + o(1) \\ & \to -\frac{t^2}{2}\,. \end{align}$$

Note that the sequence of functions is always bounded by the following integrable function,

$$\begin{equation} f(t) = \begin{cases} e^{-t^2/2} & t < 0;\\ (1+t)e^{-t} & t\ge 0. \end{cases} \end{equation}$$

The left part is easy to check by the Taylor series. The right part can be verified calculating the derivative of the difference $\ln f - \ln f_n$. Then, by the dominant convergence theorem, we have

$$\begin{equation} \int_{-\sqrt{n}}^\infty \left(1+\frac{t}{\sqrt{n}}\right)^n e^{-t\sqrt{n}}\operatorname{d}t \to \int_{-\infty}^\infty e^{-t^2/2}\operatorname{d}t = \sqrt{2\pi}\,. \end{equation}$$