# Wallis' formula Theorem: $\frac{\pi}{2}=\prod_{n=1}^\infty \frac{(2n)(2n)}{(2n-1)(2n+1)}=\frac{2\cdot2}{1\cdot3}\cdot\frac{4\cdot 4}{3\cdot 5}\cdot\frac{6\cdot 6}{5\cdot 7}\cdots=\prod_{n=1}^\infty\frac{4n^2}{4n^2-1}$ Proof: define $I_n=\int_0^{\pi/2}\sin^n(x)dx$. Using integration by part, $I_n=\frac{n-1}{n}I_{n-2}$ and $I_0=\frac{\pi}{2},I_1=1$. Use the fact that on $[0,\frac{\pi}{2}]$ , $\sin^{2n+1}(x)\le\sin^{2n}(x)\le\sin^{2n-1}(x)$ Therefore, $I_{2n+1}\le I_{2n}\le I_{2n-1}$. we have, $I_{2n}=\frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdots \frac{1}{2}\cdot\frac{\pi}{2}$ , $I_{2n+1}=\frac{2n}{2n+1}\cdot\frac{2n-2}{2n-1}\cdots\frac{2}{3}$ and $I_{2n-1}=\frac{2n-2}{2n-1}\cdots\frac{2}{3}$ $ \frac{2n}{2n+1}\cdot\frac{2n-2}{2n-1}\cdots\frac{2}{3}\le \frac{2n-1}{2n}\cdot\frac{2n-3}{2n-2}\cdots \frac{1}{2}\cdot\frac{\pi}{2}\le\frac{2n-2}{2n-1}\cdots\frac{2}{3} $ rearranging the inequality: $ \prod_{k=1}^n\frac{(2k)^2}{(2k-1)(2k+1)} \le \frac{\pi}{2}\le \frac{2n+1}{2n}\prod_{k=1}^n\frac{(2k)^2}{(2k-1)(2k+1)} $ define $P_n=\prod_{k=1}^n\frac{(2k)^2}{(2k-1)(2k+1)}$ and apply to the inequalities: $P_n\le\frac{\pi}{2}\le\frac{2n+1}{2n}P_n$ therefore, $P_n\le\frac{\pi}{2}$ and $P_n\ge \frac{\pi}{2}\cdot\frac{2n}{2n+1}$ and we have $P_n\rightarrow\frac{\pi}{2}$ $\square$ ## Created 2026-06-12 19:18