Wallis product
= Wallis product
{c}
{title2=$\pi/2=\prod_{j\geq1}\frac{4j^2}{4j^2-1}$}
{wiki}
The <Wallis product> is
$$
\frac\pi2=\prod_{j=1}^{\infty}\frac{(2j)^2}{(2j-1)(2j+1)}.
$$
Its finite product is $W_n=2^{4n}(n!)^4/[(2n+1)((2n)!)^2]$. The <Wallis integrals> give $2W_n/\pi=I_{2n+1}/I_{2n}\to1$, proving both convergence of the product and its value. The strict integral bounds also give $\pi n/(2n+1)<W_n<\pi/2$.