Wallis product ($\pi /2={\prod }_{j\ge 1}\frac{4j^2}{4j^2-1}$)

The Wallis product isIts finite product is $W_n=2^{4n}(n!{)}^4/[(2n+1)((2n)!{)}^2]$. The Wallis integrals give $2W_n/\pi =I_{2n+1}/I_{2n}\to 1$, proving both convergence of the product and its value. The strict integral bounds also give $\pi n/(2n+1)<W_n<\pi /2$.