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} $$

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.