Wallis's Product/Proof 1

Theorem

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

Proof

From the Euler Formula for the sine function:

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

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

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

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \frac{2}{\pi}\)	\(=\)	\(\displaystyle \prod_{n=1}^{\infty} \left(1 - \frac{1}{4n^2}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \frac{\pi}{2}\)	\(=\)	\(\displaystyle \prod_{n=1}^{\infty} \left(\frac{4n^2}{4n^2 - 1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \prod_{n=1}^{\infty} \frac{(2n)(2n)}{(2n-1)(2n+1)}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$