Stirling's Formula/Proof 2/Lemma 4

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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 $n \to \infty$

$\blacksquare$


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