Talk:Wallis' integrals

The limit of $(W_{n})$ is zero, since ${\sqrt {n}}W_{n}\to \pi /2$. $W_{n}/W_{n+1}$ or ${\sqrt {n}}W_{n}$ is nonzero but $W_{n}\to 0$

Yes, you're right. Dan (talk) 15:35, 25 September 2015 (UTC)