The Wallis Product
\[\pi = 2{\left( \frac{2}{1} \times \frac{2}{3} \times \frac{4}{3}
\times \frac{4}{5} \times \frac{6}{5} \times ...\right) }\]
Proof: Let \(a_n = \int_0^{\pi/2} (\sin x)^n dx\).
Since \(0 \le \sin x \le 1\) for \(0 \le x \le \pi /2\) it follows that \(a_n\) is a decreasing sequence.
Integrating by parts,
\[\array{
a_n &=& -(\sin x)^{n-1}\cos x|_0^{\pi/2} &+& \int_0^{\pi/2}(n-1)(\sin x)^{n-2}
\cos^2 x dx\\
&=& 0 &+& \int_0^{\pi/2}(n-1)(\sin x)^{n-2}(1-\sin^2 x)dx
}\]
Thus \(a_n = (n-1)a_{n-2} - (n-1)a_n\), that is \(a_n = \frac{n-1}{n}a_{n-2}\).
We have that \(a_0 = \pi/2\) and \(a_1 = 1\). Applying the recurrence relation yields
\[a_{2n} = \frac{\pi}{2}\times\frac{1}{2}\times\frac{3}{4}
\times...\times\frac{2n-1}{2n}\]
and
\[a_{2n+1} = \frac{2}{3}\times\frac{4}{5}
\times...\times\frac{2n}{2n+1}\]
Since \(a_n\) is a decreasing sequence,
\[1 \le \frac{a_{2n}}{a_{2n+1}} \le \frac{a_{2n-1}}{a_{2n+1}} = 1 + \frac{1}{2n}\]
Therefore as \(n \rightarrow \infty\), the ratio \(\frac{a_{2n}}{a_{2n+1}} \rightarrow 1\), hence
\[\frac{\pi}{2}\times\frac{1\cdot 3\cdot 5 \cdots (2n-1)}
{2\cdot 4\cdot 6\cdots (2n)}\times
\frac{3\cdot 5\cdot 7 \cdots(2n+1)}
{2\cdot 4\cdot 6\cdots (2n)} \rightarrow 1\]