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 $n \to \infty$ |
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.4 \ (1)$