## 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

$\frac{\pi}{2}\times\frac{1\cdot 3\cdot 5 \cdots (2n-1)} {2\cdot 4\cdot 6\cdots (2n)}\times \frac{3\cdot 5\cdot 7 \cdots(2n+1)} {2\cdot 4\cdot 6\cdots (2n)} \rightarrow 1$

Ben Lynn blynn@cs.stanford.edu 💡