A simple derivation of the Stirling's formula

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Theorem 1. We have

\begin{equation*} n! \sim \sqrt{2\pi n} \, n^{n} e^{-n}. \tag{1} \end{equation*}

 

For the proof of this theorem, the following observations are quite useful.

 

Lemma 2. (i) For $u \geq 0$, let $f_n (u)$ by

\begin{equation*} f_n (u) = \left(1 + \frac{u}{\sqrt{n}} \right)^{n} e^{-\sqrt{n} u}, \quad n = 1, 2, 3, \cdots. \tag{2} \end{equation*}

Then $f_{n+1}(u) \leq f_n (u)$ for all $n$ and $f_n (u) \to e^{-u^2 / 2}$ pointwise.

 

(ii) For $0 \leq u \leq \sqrt{n}$, let $g_n (u)$ by

\begin{equation*} g_n (u) = \left(1 - \frac{u}{\sqrt{n}} \right)^{n} e^{\sqrt{n} u}, \quad n = 1, 2, 3, \cdots. \tag{3} \end{equation*}

Then $g_n(u) \leq e^{-u^2 / 2}$ for all $n$ and $g_n (u) \to e^{-u^2 / 2}$ pointwise.

 

The following shows the graph of the first 10 functions $y = f_1 (u)$, $\cdots$, $y = f_{10} (u)$, colored from white to black as $n$ increases.

Proof of Lemma. Let us consider $f_n (u)$ as a function $f(n, u)$ of two real variables $n > 0$ and $u > 0$. Then it is easy to check that

\begin{align*} \frac{\partial^2}{\partial n^2} \log f(n, u) = \frac{\partial^2}{\partial n^2} \left[ n \log \left( 1 + \frac{u}{\sqrt{n}} \right) - \sqrt{n} u \right] = \frac{u^3}{4 n^{3/2} \left( \sqrt{n} + u \right)^2} > 0, \end{align*}

so that $(\partial / \partial n) f(n, u)$ is increasing with respect to $n$. But since

\begin{align*} \frac{\partial}{\partial n} f(n, u) = \log \left( 1 + \frac{u}{\sqrt{n}} \right)-\frac{u \left( u + 2 \sqrt{n} \right)}{2 \sqrt{n} \left( u + \sqrt{n} \right)} \xrightarrow[]{n\to\infty} 0, \end{align*}

it follows that $(\partial / \partial n) f(n, u) < 0$. Thus $f(n, u)$ is decreasing with respect to $n$ and the first assertion follows. The second assertion is much easier and follows from the Taylor series

\begin{align*} \log g_n (u) = n \log \left(1 - \frac{u}{\sqrt{n}} \right) + \sqrt{n} u = -\frac{u^2}{2} - \frac{u^3}{3\sqrt{n}} - \frac{u^4}{4n} - \cdots. \end{align*}

 

Proof of Theorem. From a simple calculation, we have

\begin{align*} \frac{n! e^n}{n^{n} \sqrt{n}} &= \frac{1}{\sqrt{n}} \int_{0}^{\infty} \left( \frac{t}{n} \right)^{n} e^{-(t-n)} \, \mathrm{d}t \\ &= \frac{1}{\sqrt{n}} \int_{-n}^{\infty} \left( 1 + \frac{t}{n} \right)^{n} e^{-t} \, \mathrm{d}t \\ &= \frac{1}{\sqrt{n}} \int_{0}^{n} \left( 1 - \frac{t}{n} \right)^{n} e^{t} \, \mathrm{d}t + \frac{1}{\sqrt{n}} \int_{0}^{\infty} \left( 1 + \frac{t}{n} \right)^{n} e^{-t} \, \mathrm{d}t \\ &= \int_{0}^{\sqrt{n}} \left( 1 - \frac{u}{\sqrt{n}} \right)^{n} e^{\sqrt{n} u} \, \mathrm{d}u + \int_{0}^{\infty} \left( 1 + \frac{u}{\sqrt{n}} \right)^{n} e^{-\sqrt{n} u} \, \mathrm{d}u \\ &= \int_{0}^{\sqrt{n}} g_n (u) \, \mathrm{d}u + \int_{0}^{\infty} f_n (u) \, \mathrm{d}u \end{align*}

Now the lemma shows that these integrands are dominated by some integrable functions, and thus we can take the limit inside the integral. Therefore we have

\begin{align*} \lim_{n\to\infty} \frac{n! e^n}{n^{n} \sqrt{n}} &= 2\int_{0}^{\infty} e^{-u^2 / 2} \, \mathrm{d}u = \sqrt{2 \pi}. \end{align*}