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The Wallis' product for π, written down in 1655 by John Wallis, states that

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

## Proof

First of all, consider the root of sin(x)/x is ±nπ, where n = 1, 2, 3, ... Then, we can express sine as an infinite product of linear factors given by its roots:

$\frac{\sin(x)}{x} = k \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \qquad \textrm{where}~k~\textrm{is~a~constant}$

To find the constant k, taking limit on both sides:

$\lim_{x \to 0} \frac{\sin(x)}{x} = \lim_{x \to 0} \left( k \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots \right) = k$

Using the fact that:

$\lim_{x \to 0} \frac{\sin(x)}{x} = 1$ (proof)

we get k=1. Then, we obtain the Euler-Wallis formula for sine:

$\frac{\sin(x)}{x} = \left(1 - \frac{x}{\pi}\right)\left(1 + \frac{x}{\pi}\right)\left(1 - \frac{x}{2\pi}\right)\left(1 + \frac{x}{2\pi}\right)\left(1 - \frac{x}{3\pi}\right)\left(1 + \frac{x}{3\pi}\right) \cdots$
$\frac{\sin(x)}{x} = \left(1 - \frac{x^2}{\pi^2}\right)\left(1 - \frac{x^2}{4\pi^2}\right)\left(1 - \frac{x^2}{9\pi^2}\right) \cdots$

Put x=π/2,

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