# Wallis product

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

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

**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 \(\pi/2\) 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.

**Proof 1**

Let \(I(n) = \int_0^\pi \sin^nx\, dx\) (a form of Wallis integrals). Then integrating by parts gives

\[\begin{align} u &= \sin^{n-1}x \\ \Rightarrow du &= (n-1) \sin^{n-2}x \cos x\, dx \\ dv &= \sin x\, dx \\ \Rightarrow v &= -\cos x, \end{align}\]

which implies

\[\begin{align} I(n) &= \int_0^\pi \sin^nx\, dx\\ &=\int_0^\pi u\, dv \\ &= uv |_{x=0}^{x=\pi}-\int_0^\pi v\, du \\ &= -\sin^{n-1}x\cos x |_{x=0}^{x=\pi} - \int_0^\pi - \cos x(n-1) \sin^{n-2}x \cos x\, dx \\ &= 0 - (n-1) \int_0^\pi -\cos^2x \sin^{n-2}x\, dx \qquad (n > 1) \\ &= (n - 1) \int_0^\pi \left(1-\sin^2 x\right) \sin^{n-2}x\, dx \\ &= (n - 1) \int_0^\pi \sin^{n-2}x\, dx - (n - 1) \int_0^\pi \sin^{n}x\, dx \\ &= (n - 1) I(n-2)-(n-1) I(n) \\ &= \frac{n-1}{n} I(n-2) \\ \Rightarrow \frac{I(n)}{I(n-2)} &= \frac{n-1}{n} \\ \Rightarrow \frac{I(2n-1)}{I(2n+1)} &=\frac{2n+1}{2n}. \end{align}\]

This result will be used below:

\[\begin{align} I(0) &= \int_0^\pi\, dx = x|_0^\pi = \pi \\ I(1) &= \int_0^\pi \sin x\, dx = -\cos x|_0^\pi = (-\cos \pi)-(-\cos 0) = -(-1)-(-1) = 2 \\ I(2n) &= \int_0^\pi \sin^{2n}x\, dx = \frac{2n-1}{2n}I(2n-2) = \frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2}I(2n-4). \end{align}\]

Repeating the process gives

\[\begin{align} \frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2} \cdot \frac{2n-5}{2n-4} \cdots \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} I(0) &=\pi \prod_{k=1}^n \frac{2k-1}{2k}I(2n+1)\\ &=\int_0^\pi \sin^{2n+1}x\, dx\\ &=\frac{2n}{2n+1}I(2n-1)\\ &=\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1}I(2n-3). \end{align}\]

Repeating this process again gives

\[\begin{align} \frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1} \cdot \frac{2n-4}{2n-3} \cdots \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} I(1) &=2 \prod_{k=1}^n \frac{2k}{2k+1}\sin^{2n+1}x \\ &\le \sin^{2n}x \\ &\le \sin^{2n-1}x \qquad (0 \le x \le \pi)\\ \Rightarrow I(2n+1) &\le I(2n) \le I(2n-1)\\ \Rightarrow 1 &\le \frac{I(2n)}{I(2n+1)} \le \frac{I(2n-1)}{I(2n+1)}=\frac{2n+1}{2n} \end{align}\]

from the above results.

Then by the squeeze theorem,

\[\begin{align} \lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}&=1\\ \lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}&=\frac{\pi}{2} \lim_{n\rightarrow\infty} \prod_{k=1}^n \left(\frac{2k-1}{2k} \cdot \frac{2k+1}{2k}\right)\\ &=1\\ \Rightarrow \frac{\pi}{2} &=\prod_{k=1}^\infty \left(\frac{2k}{2k-1} \cdot \frac{2k}{2k+1}\right)\\ &=\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots. \end{align}\]

Hence we are done by comparison of the integral. \(_\square\)

**Proof 2**

However Euler's infinite product, with the help of which he found the solution to the Basel problem, provides a shorter proof as follows:

\[ \frac{\sin x}{x} = \prod_{n=1}^\infty\left(1 - \frac{x^2}{n^2\pi^2}\right).\]

Let \(x = \frac{\pi}{2},\) then

\[\begin{align} \frac{2}{\pi} &= \prod_{n=1}^{\infty} \left(1 - \frac{1}{4n^2}\right) \\ \Rightarrow\frac{\pi}{2} &= \prod_{n=1}^{\infty} \left(\frac{4n^2}{4n^2 - 1}\right) \\ &= \prod_{n=1}^{\infty} \left(\frac{2n}{2n-1}\cdot\frac{2n}{2n+1}\right) \\ &= \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdots. \ _\square \end{align} \]