# Wallis product

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**Wallis product** can be stated as the infinite product

$\begin{aligned} \dfrac\pi 2 &= \prod_{n=1}^\infty \left ( \dfrac{2n}{2n-1} \cdot \dfrac{2n}{2n+1} \right) \\ &= \dfrac21 \times \dfrac23 \times \dfrac43 \times \dfrac 45 \times\dfrac65 \times \dfrac67 \times \dfrac87 \times \dfrac89 \times \cdots . \end{aligned}$

**Proofs:**

The tantalizing statement can be proved by the application of integrals, which is quite unexpected, yet an approach that is quite logical to follow by observing the $\frac{\pi}2$ on one side.

That in fact is the exact method Wallis used to prove it, by integration, by comparing $\int_0^\pi \sin^nx \, dx$ for even and odd values of $n,$ and using the fact that increasing $n$ by 1 results in a change that decreases the change in $n$ as $n$ gets larger and larger.

This proof does not require integration (see this problem Wallis Product Proof).