Proofs: Stirling's approximation

The sharp form of Stirling's formula, with the constant and the error bounds full

This proof shows: stirling-formula.

Statement. There is a real sequence $(r_n)_{n \ge 1}$ with $0 < r_n < \frac{1}{12n}$ for each $n \ge 1$, such that

In particular $n! \sim \sqrt{2\pi n}\,(n/e)^n$. The sharper lower bound $r_n > \frac{1}{12n+1}$ is Robbins' theorem and is not proved here.

Step 1: a convergent sequence. For $n \ge 1$ let

Then

To see this, expand $a_{n+1}$ and use $\ln (n+1)! = \ln n! + \ln(n+1)$: the terms $\ln n!$ cancel, and what remains is $-\ln(n+1) + (n + \frac32)\ln(n+1) - (n+\frac12)\ln n + n - (n+1)$, which is the right side.

Step 2: a series for the difference. Put $t = \frac{1}{2n+1}$, so $0 < t < 1$. Then $\frac{1+t}{1-t} = \frac{2n+2}{2n} = \frac{n+1}{n}$ and $n + \frac12 = \frac{1}{2t}$. For $0 \le t < 1$ we have $\ln\frac{1+t}{1-t} = 2\sum_{j \ge 0} \frac{t^{2j+1}}{2j+1}$. So

Step 3: two-sided bound on the difference. Every term of the series is positive, so $a_n - a_{n+1} > 0$. Also $\frac{1}{2j+1} \le \frac13$ for $j \ge 1$, with strict inequality for $j \ge 2$, so

Since $t^{-2} - 1 = (2n+1)^2 - 1 = 4n(n+1)$, we get $\frac{t^2}{1-t^2} = \frac{1}{4n(n+1)}$. Hence

Step 4: the limit and the bounds on $r_n$. Step 3 says two things. First, $(a_n)$ is strictly decreasing. Second, $a_n - \frac{1}{12n} < a_{n+1} - \frac{1}{12(n+1)}$, so $b_n = a_n - \frac{1}{12n}$ is strictly increasing. Also $b_n < a_n \le a_1$ for each $n$, so $(b_n)$ is increasing and bounded above. It converges to a limit $L$. Since $a_n = b_n + \frac{1}{12n}$ and $\frac1{12n} \to 0$, the sequence $(a_n)$ also converges to $L$. A strictly increasing sequence is smaller than its limit, and a strictly decreasing sequence is larger than its limit. So for each $n \ge 1$,

Let $r_n = a_n - L$. Then $0 < r_n < \frac{1}{12n}$, and $n! = e^{L} e^{r_n}\sqrt n\,(n/e)^n$. Put $K = e^L$. It remains to show $K = \sqrt{2\pi}$.

Step 5: Wallis integrals. For integers $m \ge 0$ let $I_m = \int_0^{\pi/2}\sin^m x\,dx$. Then $I_0 = \pi/2$ and $I_1 = 1$. For $m \ge 2$ write $\sin^m x = \sin^{m-1}x\cdot\sin x$ and integrate by parts: $I_m = (m-1)\int_0^{\pi/2}\sin^{m-2}x\cos^2x\,dx = (m-1)(I_{m-2} - I_m)$, because the boundary term $-\sin^{m-1}x\cos x$ is 0 at both ends. So $I_m = \frac{m-1}{m}I_{m-2}$. Since $0 \le \sin x \le 1$ on $[0, \pi/2]$, the sequence $(I_m)$ is non-increasing. By induction on $m$, $I_m I_{m+1} = \frac{\pi}{2(m+1)}$: for $m = 0$ the product is $\pi/2$, and if it holds for $m$, then $I_{m+1}I_{m+2} = \frac{m+1}{m+2}I_{m+1}I_m = \frac{\pi}{2(m+2)}$.

For $n \ge 1$ the monotonicity gives $I_{2n+1} \le I_{2n} \le I_{2n-1}$. Multiply by $I_{2n} > 0$:

Multiply by $4n/\pi$: the left side becomes $\frac{2n}{2n+1}$ and the right side becomes 1. Both tend to 1, so $I_{2n}^2 \sim \frac{\pi}{4n}$, that is $I_{2n} \sim \frac12\sqrt{\pi/n}$.

Step 6: the central binomial coefficient. The recurrence gives $I_{2n} = \frac\pi2\prod_{j=1}^{n}\frac{2j-1}{2j}$. Now $\prod_{j=1}^n (2j-1) = \frac{(2n)!}{2^n n!}$ and $\prod_{j=1}^n 2j = 2^n n!$, so $I_{2n} = \frac{\pi}{2}\cdot\frac{(2n)!}{4^n (n!)^2} = \frac\pi2\binom{2n}{n}4^{-n}$. With Step 5,

Step 7: the constant. By Step 4, $n! \sim K\sqrt n\,(n/e)^n$. So $(2n)! \sim K\sqrt{2n}\,(2n/e)^{2n}$ and $(n!)^2 \sim K^2 n\,(n/e)^{2n}$. Dividing,

Compare with Step 6: $\frac{\sqrt 2}{K\sqrt n} \sim \frac{1}{\sqrt{\pi n}}$, so $K = \sqrt{2\pi}$. Then $n! = \sqrt{2\pi n}\,(n/e)^n e^{r_n}$ with $0 < r_n < \frac1{12n}$, and $e^{r_n} \to 1$. $\blacksquare$

Remark. The lower bound $r_n > \frac1{12n+1}$ needs a finer estimate of the series of Step 2. Robbins gives it in the paper cited on the page. The proof above gives the upper bound $\frac{1}{12n}$ and the positivity of $r_n$.