# Wallis product

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

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

## Derivation

Wallis derived this infinite product as it is done in calculus books today, by comparing $\scriptstyle \int_0^\pi \sin^nx dx$ for even and odd values of n, and noting that for large n, increasing n by 1 results in a change that becomes ever smaller as n increases. Since infinitesimal calculus as we know it did not yet exist then, and the mathematical analysis of the time was inadequate to discuss the convergence issues, this was a hard piece of research, and tentative as well.

Wallis' product is, in retrospect, an easy corollary of the later Euler formula for the sine function.

## Proof using Euler's infinite product for the sine function[1]

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

Let x = π2:

\begin{align} \Rightarrow\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 \end{align}

## Proof using integration[2]

Let:

$I(n) = \int_0^\pi \sin^nxdx$

(a form of Wallis' integrals). Integrate by parts:

\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}
\begin{align} \Rightarrow I(n) &= \int_0^\pi \sin^nxdx=\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, n > 1 \\ {} &= (n - 1) \int_0^\pi (1-\sin^2 x) \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 xdx = -\cos x|_0^\pi = (-\cos \pi)-(-\cos 0) = -(-1)-(-1) = 2 \\ I(2n) &= \int_0^\pi \sin^{2n}xdx = \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,

$=\frac{2n-1}{2n} \cdot \frac{2n-3}{2n-2} \cdot \frac{2n-5}{2n-4} \cdot \cdots \cdot \frac{5}{6} \cdot \frac{3}{4} \cdot \frac{1}{2} I(0)=\pi \prod_{k=1}^n \frac{2k-1}{2k}$
$I(2n+1)=\int_0^\pi \sin^{2n+1}xdx=\frac{2n}{2n+1}I(2n-1)=\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1}I(2n-3)$

Repeating the process,

$=\frac{2n}{2n+1} \cdot \frac{2n-2}{2n-1} \cdot \frac{2n-4}{2n-3} \cdot \cdots \cdot \frac{6}{7} \cdot \frac{4}{5} \cdot \frac{2}{3} I(1)=2 \prod_{k=1}^n \frac{2k}{2k+1}$
$\sin^{2n+1}x \le \sin^{2n}x \le \sin^{2n-1}x, 0 \le x \le \pi$
$\Rightarrow I(2n+1) \le I(2n) \le I(2n-1)$
$\Rightarrow 1 \le \frac{I(2n)}{I(2n+1)} \le \frac{I(2n-1)}{I(2n+1)}=\frac{2n+1}{2n}$, from above results.

By the squeeze theorem,

$\Rightarrow \lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}=1$
$\lim_{n\rightarrow\infty} \frac{I(2n)}{I(2n+1)}=\frac{\pi}{2} \lim_{n\rightarrow\infty} \prod_{k=1}^n \left(\frac{2k-1}{2k} \cdot \frac{2k+1}{2k}\right)=1$
$\Rightarrow \frac{\pi}{2}=\prod_{k=1}^\infty \left(\frac{2k}{2k-1} \cdot \frac{2k}{2k+1}\right)=\frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \cdots$

## Relation to Stirling's approximation

Stirling's approximation for n! asserts that

$n! = \sqrt {2\pi n} {\left(\frac{n}{e}\right)}^n \left[1 + O\left(\frac{1}{n}\right) \right]$

as n → ∞. Consider now the finite approximations to the Wallis product, obtained by taking the first k terms in the product:

$p_k = \prod_{n=1}^{k} \frac{2n}{2n - 1}\frac{2n}{2n + 1}$

pk can be written as

\begin{align} p_k &= {1 \over {2k + 1}} \prod_{n=1}^{k} \frac{(2n)^4}{[(2n)(2n - 1)]^2} \\ &= {1 \over {2k + 1}} \cdot {{2^{4k}\,(k!)^4} \over {[(2k)!]^2}} \end{align}

Substituting Stirling's approximation in this expression (both for k! and (2k)!) one can deduce (after a short calculation) that pk converges to π2 as k → ∞.

## ζ'(0)[1]

The Riemann zeta function and the Dirichlet eta function can be defined:

\begin{align} \zeta(s) &= \sum_{n=1}^\infty \frac{1}{n^s}, \Re(s)>1 \\ \eta(s) &= (1-2^{1-s})\zeta(s) \\ &= \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}, \Re(s)>0 \end{align}

Applying an Euler transform to the latter series, the following is obtained:

\begin{align} \eta(s) &= \frac{1}{2}+\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\frac{1}{n^s}-\frac{1}{(n+1)^s}\right], \Re(s)>-1 \\ \Rightarrow \eta'(s) &= (1-2^{1-s})\zeta'(s)+2^{1-s} (\ln 2) \zeta(s) \\ &= -\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\frac{\ln n}{n^s}-\frac{\ln (n+1)}{(n+1)^s}\right], \Re(s)>-1 \end{align}
\begin{align} \Rightarrow \eta'(0) &= -\zeta'(0) - \ln 2 = -\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\left[\ln n-\ln (n+1)\right] \\ &= -\frac{1}{2} \sum_{n=1}^\infty (-1)^{n-1}\ln \frac{n}{n+1} \\ &= -\frac{1}{2} \left(\ln \frac{1}{2} - \ln \frac{2}{3} + \ln \frac{3}{4} - \ln \frac{4}{5} + \ln \frac{5}{6} - \cdots\right) \\ &= \frac{1}{2} \left(\ln \frac{2}{1} + \ln \frac{2}{3} + \ln \frac{4}{3} + \ln \frac{4}{5} + \ln \frac{6}{5} + \cdots\right) \\ &= \frac{1}{2} \ln\left(\frac{2}{1}\cdot\frac{2}{3}\cdot\frac{4}{3}\cdot\frac{4}{5}\cdot\cdots\right) = \frac{1}{2} \ln\frac{\pi}{2} \\ \Rightarrow \zeta'(0) &= -\frac{1}{2} \ln\left(2 \pi\right) \end{align}