# Wallis's Product/Proof 1

## Theorem

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

## Proof

 $\displaystyle \dfrac {\sin x} x$ $=$ $\displaystyle \paren {1 - \dfrac {x^2} {\pi^2} } \paren {1 - \dfrac {x^2} {4 \pi^2} } \paren {1 - \dfrac {x^2} {9 \pi^2} } \cdots$ $\displaystyle$ $=$ $\displaystyle \prod_{n \mathop = 1}^\infty \paren {1 - \dfrac {x^2} {n^2 \pi^2} }$

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

$\sin \dfrac \pi 2 = 1$

Hence:

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

$\blacksquare$

## Source of Name

This entry was named for John Wallis.