Define, for an integer $n \geq 0$,

$$I_{n}=\int_{0}^{\pi / 2} \sin ^{n} x d x$$

Show that for every $n \geq 2, n I_{n}=(n-1) I_{n-2}$, and deduce that

$$I_{2 n}=\frac{(2 n) !}{\left(2^{n} n !\right)^{2}} \frac{\pi}{2} \quad \text { and } \quad I_{2 n+1}=\frac{\left(2^{n} n !\right)^{2}}{(2 n+1) !}$$

Show that $0<I_{n}<I_{n-1}$, and that

$$\frac{2 n}{2 n+1}<\frac{I_{2 n+1}}{I_{2 n}}<1$$

Hence prove that

$$\lim _{n \rightarrow \infty} \frac{2^{4 n+1}(n !)^{4}}{(2 n+1)(2 n) !^{2}}=\pi .$$