Analysis | Part IA, 2004

Define, for an integer n0n \geq 0,

In=0π/2sinnxdxI_{n}=\int_{0}^{\pi / 2} \sin ^{n} x d x

Show that for every n2,nIn=(n1)In2n \geq 2, n I_{n}=(n-1) I_{n-2}, and deduce that

I2n=(2n)!(2nn!)2π2 and I2n+1=(2nn!)2(2n+1)!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<In<In10<I_{n}<I_{n-1}, and that

2n2n+1<I2n+1I2n<1\frac{2 n}{2 n+1}<\frac{I_{2 n+1}}{I_{2 n}}<1

Hence prove that

limn24n+1(n!)4(2n+1)(2n)!2=π.\lim _{n \rightarrow \infty} \frac{2^{4 n+1}(n !)^{4}}{(2 n+1)(2 n) !^{2}}=\pi .

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