Asymptotic behavior of Wallis integrals

Let us denote with $W_n$ the Wallis integral, that is:

$$W_n=\int_0^{\pi/2} \sin^n x \, {\rm d}x$$

Prove that $W_n \sim \sqrt{\frac{\pi}{2n}}$.

Solution

It is well known that:

$$W_n= \frac{\sqrt{\pi}}{2}\frac{\Gamma \left ( \frac{n+1}{2} \right )}{\Gamma \left ( \frac{n}{2}+1 \right )}$$

where $\Gamma$ is the Euler's Gamma function. Now,

\begin{align*}
W_n W_{n+1} &=\frac{\sqrt{\pi}}{2}\frac{\Gamma \left ( \frac{n+1}{2} \right )}{\Gamma \left ( \frac{n}{2}+1 \right )} \frac{\sqrt{\pi}}{2} \frac{\Gamma \left ( \frac{n}{2}+1 \right )}{\Gamma \left ( \frac{n+1}{2}+ 1  \right )} \\
 &=\frac{\pi}{4} \frac{\Gamma \left ( \frac{n+1}{2} \right )}{\Gamma \left ( \frac{n+1}{2}+ 1  \right )} \\
 &= \frac{\pi}{4} \frac{\Gamma \left ( \frac{n+1}{2} \right )}{\frac{n+1}{2} \Gamma \left ( \frac{n+1}{2} \right )}\\
 &=\frac{\pi}{2 (n+1)} \\
 &\sim \frac{\pi}{2n }
\end{align*}

Now, it is obvious that $\lim W_n =0$ as well as $\lim \sqrt{n} W_n = \sqrt{\frac{\pi}{2}}$.