Wallis's Product

Theorem


\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:


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

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

From Nature of Sine Function we note that $$\sin \frac \pi 2 = 1$$, and hence:

$$ $$ $$ $$

Wallis's Original Proof
Wallis, of course, had no recourse to Euler's techniques.

He did this job by comparing $$\int_0^\pi \sin^n x dx$$ for even and odd values of $$n$$, and noting that for large $$n$$, increasing $$n$$ by 1 makes little change.