Stirling's Formula/Proof 2/Lemma 5

Theorem

$\dfrac {n!} {n^n \sqrt n e^{-n} } \to \sqrt {2 \pi}$ as $n \to \infty$

Proof

By previous work done in Stirling's Formula: Proof 2 it is noted that $\sequence {\dfrac {n!} {n^n \sqrt n e^{-n} } }_{n \mathop \in \N}$ is a convergent sequence.

Let $\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$.

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

Then:

\(\ds \frac {I_{2 n} } {I_{2 n + 1} }\)

\(=\)

\(\ds \frac {\paren {2 n}!} {\paren {2^n n!}^2} \frac \pi 2 \cdot \frac {\paren {2 n + 1}!} {\paren {2^n n!}^2}\)

Definite Integral from $0$ to $\dfrac \pi 2$ of Even Power of $\sin x$ and Corollary 2

\(\ds \)

\(=\)

\(\ds \frac \pi 2 \paren {2 n + 1} \frac {\paren {\paren {2 n}!}^2} {\paren {2^n n!}^4}\)

extracting $\paren {2 n + 1}$ as a factor and rearranging

\(\ds \)

\(\sim\)

\(\ds \frac \pi 2 \paren {2 n + 1} \frac {\paren {C \paren {2 n}^{2 n + 1/2} e^{-2 n} }^2} {\paren {2^n C n^{n + 1/2} e^{-n} }^4}\)

entering $C$ into top and bottom

\(\ds \)

\(=\)

\(\ds \frac \pi 2 \frac {\paren {2 n + 1} } {C^2} \frac {2^{4 n} 2 \paren {n^{4 n + 1} } e^{-4 n} } {2^{4 n} \paren {n^{4 n + 2} } e^{-4 n} }\)

rearranging

\(\ds \)

\(=\)

\(\ds \frac {\pi \paren {2 n + 1} } {n C^2}\)

much simplification

\(\ds \)

\(\to\)

\(\ds \frac {2 \pi} {C^2}\)

\(\ds \text {as $n \to \infty$}\)

From Lemma 4:

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

from which:

\(\ds \frac {2 \pi} {C^2}\)

\(=\)

\(\ds 1\)

\(\ds \leadsto \ \ \)

\(\ds C^2\)

\(=\)

\(\ds 2 \pi\)

Hence the result.

$\blacksquare$

Sources

• 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.4 \ (2)$