Rigorous Proof of Stirling's Approximation Formula
Theorem
$$ \lim_{n \to \infty} {{n!} \over {e^{n \ln n - n} \sqrt{ 2 \pi n} }} = 1 $$
Description
The Stirling approximation, also known as the Stirling’s formula, is widely used in various fields such as statistics and physics. It can also be expressed using the gamma function as follows: $$ \Gamma ( n ) \approx {e^{n \ln n - n} \sqrt{ 2 \pi n}} $$
This proof is included in the appendix of a book titled “The Secrets of the Zeta Function,” written by Nobushige Kurokawa, a mathematician from Japan, who discovered it during his high school years and later published it in the Mathematics Seminar June 1972 issue. However, upon reviewing it, there were many omissions, and it wasn’t much of a reference; in fact, the argument wasn’t a proof at all. Similarly, consulting “The Secrets of the Zeta Function” directly wouldn’t be much helpful. It is confidently asserted that this post explains Kurokawa’s proof in the easiest and most friendly manner available in Korea.
There are proofs in other analytical books, but the reason for choosing this particular one is that it was easier than any other proof seen so far without forsaking rigor. Of course, the actual proof, including the proving of auxiliary theorems, would be longer, but at least it was possible to avoid the futile effort of proving auxiliary theorems solely for the Stirling approximation. The auxiliary theorem introduced in this proof method comes from a well-structured context, making it quite enjoyable to follow the process as well.
See Also
Proof
Let’s define the function $f( x ) := \ln (1 + x)$.
Part 1. $\displaystyle \lim_{n \to \infty} \left[ \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx \right] = {{f(1) - f(0) } \over {2}}$
Since $\displaystyle f '(x) = {{1} \over {1 + x}}$, for $\displaystyle x \in \left[ {{k-1} \over {n}} , {{k} \over {n}} \right]$, the minimum value is $\displaystyle {{1} \over {1 + {{k} \over {n}} }}$, and the maximum value is $\displaystyle {{1} \over {1 + {{k-1} \over {n}} }}$. Therefore,
$$ {{1} \over {1 + {{k} \over {n}} }} \le {{ f \left( {{k} \over {n}} \right) - f(x) } \over { {{k} \over {n}} - x }} \le {{1} \over {1 + {{k-1} \over {n}} }} $$
Multiplying both sides by $\displaystyle \left( {{k} \over {n}} - x \right)$,
$$ {{1} \over {1 + {{k} \over {n}} }} \left( {{k} \over {n}} - x \right) \le f \left( {{k} \over {n}} \right) - f(x) \le {{1} \over {1 + {{k-1} \over {n}} }} \left( {{k} \over {n}} - x \right) $$
Integrating both sides from $\displaystyle {{k-1} \over {n}}$ to $\displaystyle {{k} \over {n}}$,
$$ \begin{align*} \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left( {{k} \over {n}} - x \right) dx =& {{k} \over {n}} {{1} \over {n}} - \left[ {{1} \over {2}} x^2 \right]_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \\ =& {{2k} \over {2n^2}} - {{k^2 - k^2 + 2k - 1} \over {2n^2}} \\ =& {{1} \over {2n^2}} \end{align*} $$
Therefore,
$$ {{1} \over {2n^2}} {{1} \over {1 + {{k} \over {n}} }} \le \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left[ f \left( {{k} \over {n}} \right) - f(x) \right] dx \le {{1} \over {2n^2}} {{1} \over {1 + {{k-1} \over {n}} }} $$
Multiplying both sides by $n$,
$$ {{1} \over {2n}} {{1} \over {1 + {{k} \over {n}} }} \le n \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left[ f \left( {{k} \over {n}} \right) - f(x) \right] dx \le {{1} \over {2n}} {{1} \over {1 + {{k-1} \over {n}} }} $$
Adding both sides from $k=1$ to $n$,
$$ {{1} \over {2}} \sum_{k=1}^{n} {{1} \over {n}} {{1} \over {1 + {{k} \over {n}} }} \le n \sum_{k=1}^{n} \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left[ f \left( {{k} \over {n}} \right) - f(x) \right] dx \le {{1} \over {2}} \sum_{k=1}^{n} {{1} \over {n}} {{1} \over {1 + {{k-1} \over {n}} }} $$
The middle term is,
$$ \begin{align*} n \sum_{k=1}^{n} \int_{ {{k-1} \over {n}} }^{ {{k} \over {n}} } \left[ f \left( {{k} \over {n}} \right) - f(x) \right] dx =& n \left[ {{k} \over {n}} - {{k-1} \over {n}} \right] \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx \\ =& \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx \end{align*} $$
Riemann Sum and Definite Integral: $$\lim _{ n\to \infty }{ \sum _{ k=1 }^{ n }{ f\left( a+\frac { p }{ n }k \right) \frac { p }{ n } } } =\int _{ 0 }^{ 1 }{ pf(a+px)dx }$$
The left side and the right side, when $n \to \infty$,
$$ \begin{align*} {{1} \over {2}} \sum_{k=1}^{\infty} {{1} \over {n}} {{1} \over {1 + {{k} \over {n}} }} =& {{1} \over {2}} \sum_{k=1}^{\infty} {{1} \over {n}} {{1} \over {1 + {{k-1} \over {n}} }} \\ =& {{1} \over {2}} \int_{0}^{1} {{1} \over {1 + x}} dx \\ =& {{1} \over {2}}\left[ \ln | 1 + x | \right]_{0}^{1} \\ =& {{f(1) - f(0) } \over {2}} \end{align*} $$
According to the Sandwich Theorem,
$$ \lim_{n \to \infty} \left[ \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx \right] = {{f(1) - f(0) } \over {2}} $$
Part 2. $\displaystyle \lim_{n \to \infty} \left[ {{(2n)!} \over {n! n^n}} \left( {{e} \over {4}} \right)^{n} \right] = \sqrt{2}$
Looking at each term obtained from Part 1, The first term $\displaystyle \sum_{k=1}^{n} f \left( {{k} \over {n}} \right)$ is,
$$ \begin{align*} \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) =& \sum_{k=1}^{n} \ln \left( {{n+k} \over {n}} \right) \\ =& \ln \left( {{n+1} \over {n}} {{n+2} \over {n}} \cdots {{n+n} \over {n}} \right) \\ =& \ln \left( {{ (2n)! } \over { n! n^n }} \right) \end{align*} $$
The second term $\displaystyle n \int_{0}^{1} f(x) dx$ is,
$$ \begin{align*} n \int_{0}^{1} f(x) dx =& n \left[ (1+x) \ln (1+x) -x \right]_{0}^{1} \\ =& 2n \ln 2 - n \\ =& \ln \left( {{2^{2n}} \over {e^{n}}} \right) \\ =& \ln \left( {{4} \over {e}} \right)^{n} \end{align*} $$
Therefore,
$$ \begin{align*} \lim_{n \to \infty} \left[ \sum_{k=1}^{n} f \left( {{k} \over {n}} \right) - n \int_{0}^{1} f(x) dx \right] =& \lim_{n \to \infty} \ln \left[ {{(2n)!} \over {n! n^n}} \left( {{e} \over {4}} \right)^{n} \right] \\ =& {{ \ln 2 } \over {2}} \\ =& \ln \sqrt{2} \end{align*} $$
And summarizing,
$$ \lim_{n \to \infty} \left[ {{(2n)!} \over {n! n^n}} \left( {{e} \over {4}} \right)^{n} \right] = \sqrt{2} $$
Part 3. $\displaystyle \lim_{n \to \infty} {{4^n (n!)^2 } \over {\sqrt{n} (2n)! } } = \sqrt{ \pi }$
Wallis Product: $$ \prod_{n=1}^{\infty} {{4n^2} \over {4n^2 - 1}} = \lim_{n \to \infty} {{2 \cdot 2 } \over { 1 \cdot 3 } } \cdot {{4 \cdot 4 } \over { 3 \cdot 5 } } \cdot \cdots \cdot {{2n \cdot 2n } \over { (2n-1) \cdot (2n+1) } } = {{ \pi } \over {2}} $$
$$ \begin{align*} {{\pi} \over {2}} =& \lim_{n \to \infty} {{2^2 \cdot 4^2 \cdot \cdots \cdot (2n)^2} \over {1 \cdot 3^2 \cdot \cdots \cdot (2n-1)^2 \cdot (2n+1)}} \\ =& \lim_{n \to \infty} {{2^2 \cdot 4^2 \cdot \cdots \cdot (2n)^2} \over {1 \cdot 3^2 \cdot \cdots \cdot (2n-1)^2 \cdot 2n }} \end{align*} $$
Multiplying both sides by $2$,
$$ \pi = \lim_{n \to \infty} {{2^2 \cdot 4^2 \cdot \cdots \cdot (2n)^2} \over {1 \cdot 3^2 \cdot \cdots \cdot (2n-1)^2 \cdot n }} $$
Taking the square root of both sides,
$$ \begin{align*} \sqrt{ \pi } =& \lim_{n \to \infty} {{1} \over {\sqrt{n}}} {{2 \cdot 4 \cdot \cdots \cdot (2n)} \over {1 \cdot 3 \cdot \cdots \cdot (2n-1) }} \\ =& \lim_{ n \to \infty} {{1} \over {\sqrt{n}}} {{2 \cdot 4 \cdot \cdots \cdot (2n) \cdot 2 \cdot 4 \cdot \cdots \cdot (2n)} \over {1 \cdot 2 \cdot 3 \cdot \cdots \cdot (2n-1) \cdot 2n }} \\ =& \lim_{n \to \infty} {{4^n (n!)^2 } \over {\sqrt{n} (2n)! }} \end{align*} $$
Summarizing,
$$ \lim_{n \to \infty} {{4^n (n!)^2 } \over {\sqrt{n} (2n)! } } = \sqrt{ \pi } $$
Part 4. Synthesis
$$ \begin{align*} \lim_{n \to \infty} {{n!} \over {e^{n \ln n - n }\sqrt{n}} } =& \lim_{n \to \infty} {{n!} \over {n^{n} e^{-n} \sqrt{n} }} \\ =& \lim_{n \to \infty} { { (2n)!} \over { 4^{n} n! } } \left( {{e} \over {n}} \right)^{n} {{4^n (n!)^2} \over {\sqrt{ n} (2n)! }} \end{align*} $$
According to Part 2 and 3, $$ \lim_{n \to \infty} { { (2n)!} \over { 4^{n} n! } } \left( {{e} \over {n}} \right)^{n} {{4^n (n!)^2} \over {\sqrt{ n} (2n)! }} = \sqrt{2} \sqrt{\pi} = \sqrt{2 \pi} $$ Therefore, $$ \lim_{n \to \infty} {{n!} \over {e^{n \ln n - n} \sqrt{ 2 \pi n} }} = 1 $$
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