# Stirling's Formula/Proof 2/Lemma 4

## Theorem

Let $I_n$ be defined as:

$\displaystyle I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$

Then:

$\displaystyle \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$

## Proof

 $\displaystyle n I_n$ $=$ $\displaystyle \left({n - 1}\right) I_{n - 2}$ Reduction Formula for Definite Integral of Power of Sine $\displaystyle \implies \ \$ $\displaystyle 1$ $<$ $\displaystyle \frac {I_{ 2n} } {I_{2 n + 1} }$ $\displaystyle$ $<$ $\displaystyle \frac {I_{2 n - 1} } {I_{2 n + 1} }$ Reduction Formula for Definite Integral of Power of Sine $\displaystyle$ $=$ $\displaystyle \frac {2 n + 1} {2 n}$ Reduction Formula for Definite Integral of Power of Sine $\displaystyle$ $=$ $\displaystyle \to 0$ as $n \to \infty$

$\blacksquare$