# Stirling’s formula Numerical approximation of factorial. Theorem: $n!\sim\sqrt{2\pi n}(\frac{n}{e})^n$ Proof: define $b_n=\log(n!)-(n+\frac{1}{2})\log(n)+n$ . $b_n$ converge because $\begin{align} b_n-b_{n-1}&= \log(n)-(n+\frac{1}{2})\log(n)+(n-\frac{1}{2})\log(n-1)+1\\ &=-(n-\frac{1}{2})\log(n)+(n-\frac{1}{2})\log(n-1)+1 \\ &=(n-\frac{1}{2})\log(\frac{n-1}{n})+1\\ &=(n-\frac{1}{2})\log(1-\frac{1}{n})+1 \end{align} $ Using $\log(1-x)=-x-\frac{x^2}{2}-\frac{x^3}{3}+O(x^4)$ , with $x=\frac{1}{n}$ we get $b_n-b_{n-1}=(n-\frac{1}{2})(-\frac{1}{n}-\frac{1}{2n^2}-\frac{1}{3n^3})+1=-\frac{1}{12n^2}+O(\frac{1}{n^3})$ and since $\sum\frac{1}{n^2}<\infty$ . Let assume $b_n\rightarrow \log(C)$ for constant $C$. We have $C\sim e^{b_n}=\frac{n!e^n}{n^{n+\frac{1}{2}}}$ , and equivalently $n!\sim C\sqrt{n}(\frac{n}{e})^n$ . Let's compute $C$ . Using the formula above, $\binom{2n}{n}=\frac{(2n)!}{(n!)^2}\sim\frac{C\sqrt{2n}(\frac{2n}{e})^{2n}}{C^2 n(\frac{n}{e})^{2n}}=\frac{\sqrt2 \cdot2^{2n}}{C\sqrt n}=\frac{\sqrt 2\cdot 4^n}{C\sqrt n}$ [[Wallis' formula]] $\prod_{k=1}^n \frac{(2k)(2k)}{(2k-1)(2k+1)}\sim\frac{\pi}{2}$ implies $ \frac{\pi}{2}\sim\prod_{k=1}^n \frac{(2k)(2k)}{(2k-1)(2k+1)}=\frac{1}{(2n+1)}\prod_{k=1}^n \frac{(2k)(2k)}{(2k-1)(2k-1)}=\frac{1}{(2n+1)}\bigg(\prod_{k=1}^n \frac{(2k)}{(2k-1)}\bigg)^2 $ Therefore, $\prod_{k=1}^n \frac{(2k)}{(2k-1)}\sim \sqrt{n\pi}$ and we get $ \binom{2n}{n}=\frac{(2n)!}{(n!)^2}=\bigg(\frac{(2n)!}{2^n(n!)}\bigg)\bigg(\frac{4^n}{2^n(n!)}\bigg)=4^n\prod_{k=1}^n\frac{(2k-1)}{2k}\sim\frac{4^n}{\sqrt{n\pi}} $ Back to the main proof, $\frac{4^n}{\sqrt{\pi n}}\sim\binom{2n}{n}\sim\frac{\sqrt 2\cdot 4^n}{C\sqrt n} $ Therefore, $C=\sqrt{2\pi}$ and we get $\boxed{n!\sim \sqrt{2\pi n}\bigg(\frac{n}{e}\bigg)^n}$ $\square$ ## Created 2026-06-12 14:16