Stirling's Formula/Proof 2
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Theorem
The factorial function can be approximated by the formula:
- $n! \sim \sqrt {2 \pi n} \paren {\dfrac n e}^n$
where $\sim$ denotes asymptotically equal.
Proof
Consider the sequence $\sequence {d_n}$ defined as:
- $d_n = \map \ln {n!} - \paren {n + \dfrac 1 2} \ln n + n$
From Lemma 2 it is seen that $\sequence {d_n}$ is a decreasing sequence.
From Lemma 3 it is seen that the sequence:
- $\sequence {d_n - \dfrac 1 {12 n} }$
is increasing.
In particular:
- $\forall n \in \N_{>0}: d_n - \dfrac 1 {12 n} \ge d_1 = \dfrac 1 {12}$
and so $\sequence {d_n}$ is bounded below.
From the Monotone Convergence Theorem (Real Analysis), it follows that $\sequence {d_n}$ is convergent.
Let $d_n \to d$ as $n \to \infty$.
From Exponential Function is Continuous, we have:
- $\exp d_n \to \exp d$ as $n \to \infty$
Let $C = \exp d$.
Then:
- $\dfrac {n!} {n^n \sqrt n e^{-n} } \to C$ as $n \to \infty$
It remains to be shown that $C = \sqrt {2 \pi}$.
Let $I_n$ be defined as:
- $\ds I_n = \int_0^{\frac \pi 2} \sin^n x \rd x$
Then from Lemma 4:
- $\ds \lim_{n \mathop \to \infty} \frac {I_{2 n} } {I_{2 n + 1} } = 1$
So:
\(\ds \lim_{n \mathop \to \infty} \dfrac {n!} {n^n \sqrt n e^{-n} }\) | \(=\) | \(\ds \sqrt {2 \pi}\) | Lemma 5 | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \lim_{n \mathop \to \infty} \dfrac {n!} {\sqrt {2 \pi n} \paren {\frac n e}^n}\) | \(=\) | \(\ds 1\) |
Hence the result.
$\blacksquare$
Sources
- 1977: K.G. Binmore: Mathematical Analysis: A Straightforward Approach ... (previous) ... (next): $\S 17.2$