Stirling's Formula/Proof 2/Lemma 4

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Theorem

Let $I_n$ be defined as:

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

Then:

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


Proof

\(\ds n I_n\) \(=\) \(\ds \paren {n - 1} I_{n - 2}\) \(\quad\) Reduction Formula for Definite Integral of Power of Sine
\(\ds \leadsto \ \ \) \(\ds 1\) \(<\) \(\ds \frac {I_{ 2n} } {I_{2 n + 1} }\)
\(\ds \) \(<\) \(\ds \frac {I_{2 n - 1} } {I_{2 n + 1} }\) \(\quad\) Reduction Formula for Definite Integral of Power of Sine
\(\ds \) \(=\) \(\ds \frac {2 n + 1} {2 n}\) \(\quad\) Reduction Formula for Definite Integral of Power of Sine
\(\ds \) \(\to\) \(\ds 1\) as $n \to \infty$

$\blacksquare$


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