Wallis's Product/Proof 1

Theorem

 * $\displaystyle \prod_{n \mathop = 1}^{\infty} \frac{2n}{2n-1} \cdot \frac{2n}{2n+1} = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}$

Proof
From the Euler Formula for the sine function:


 * $\displaystyle \frac{\sin \left({x}\right)} x = \left ({1 - \frac{x^2}{\pi^2}}\right) \left({1 - \frac{x^2}{4 \pi^2}}\right) \left({1 - \frac{x^2}{9 \pi^2}}\right) \cdots = \prod_{n \mathop = 1}^\infty \left({1 - \frac{x^2}{n^2 \pi^2}}\right)$

we substitute $\displaystyle x = \frac \pi 2$.

From Sine of Half-Integer Multiple of Pi we note that $\sin \dfrac \pi 2 = 1$, and hence: