Wallis product

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

$\prod_{n=1}^{\infty} \left(\frac{2n}{2n-1} \cdot \frac{2n}{2n+1}\right) = \frac{2}{1} \cdot \frac{2}{3} \cdot \frac{4}{3} \cdot \frac{4}{5} \cdot \frac{6}{5} \cdot \frac{6}{7} \cdot \frac{8}{7} \cdot \frac{8}{9} \cdots = \frac{\pi}{2}$
Comparison of the convergence of the Wallis product (purple asterisks) and several historical infinite series for π. Sn is the approximation after taking n terms. Each subsequent subplot magnifies the shaded area horizontally by 10 times. (click for detail)

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}