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)