Pi

$$\pi =2(\frac{2}{1}\times \frac{2}{3}\times \frac{4}{3}\times \frac{4}{5}\times \frac{6}{5}\times ...)$$

Proof: Let $a_n={\int }_0^{\pi /2}(\sin x{)}^n\mathrm{dx}$.

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,

$$\begin{array}{ccccc}{a}_n& =& -(\sin x{)}^{n-1}\cos x{\mid }_0^{\pi /2}& +& {\int }_0^{\pi /2}(n-1)(\sin x{)}^{n-2}{\cos }^{2}x\mathrm{dx}\\ & =& 0& +& {\int }_0^{\pi /2}(n-1)(\sin x{)}^{n-2}(1-{\sin }^{2}x)\mathrm{dx}\end{array}$$

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 $n\to \mathrm{\infty }$, the ratio $\frac{a_{2n}}{a_{2n+1}}\to 1$, hence

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