The most beautiful formula in mathematics is certainly not the triviality $1 + e^{i\pi}=0$, as claimed by many tasteless mathematicians with no self-esteem, but rather *Stirling's formula*, stating that $n!\sim n^n e^{-n}\sqrt{2\pi n}$. Quantifying the error of this approximation can be done to any precision using the Euler-MacLaurin complicated formula, but it is often handy to have a simpler estimation; Herbert Robbins^{[1]} found one (original note here) which is surprisingly accurate:

The inequalities are strict and valid for every $n$. The proof of this result is given below. For future reference, we can compare (1) with the higher-order expansion:

$\frac{n!}{n^n e^{-n}\sqrt{2\pi n}} = \exp \left\lbrace \frac{1}{12n} - \frac{1}{360n^3} + \frac{1}{1260n^5} + O(\frac{1}{n^7}) \right\rbrace.$We have $\ln(n!) = \sum_{k=1}^{n-1} \ln(k+1)$. Robbins' proof consists in a subtle approximation of $\ln(k+1)$, as seen as the area of the rectangle $[k,k+1] \times [0, \ln(k+1)]$. This area is equal to:

the area $I_k$ of the points in the rectangle under the curve $\ln(x)$, that is $\int_k^{k+1}\ln(x)dx$, (in the picture, it's the pink+black zone)

plus the area $T_k$ of the triangle between the points $(k,\ln(k)), (k, \ln(k+1)), (k+1, \ln(k+1))$, (blue+black zone)

minus the area $\delta_k$ of the small black zone we counted twice, which is equal to $I_k$ minus the area of the points in the rectangle that are not in the preceding triangle.

Clearly, $T_k = (\ln(k+1) - \ln(k))/2$. We also recall that the antiderative of $\ln(x)$ is $x\ln(x) - x$. Consequently, noting $S_n = \delta_2 + \dotsb + \delta_{n-1}$,

$\begin{aligned} \ln(n!) &= I_2 + \dotsb + I_{n-1} + T_2 + \dotsc + T_{n-1} - S_n\\ &= \int_1^n \ln(x)dx + \frac{1}{2}\ln(k) - S_n\\ &= n\ln(n) - n + 1 + \frac{1}{2}\ln(n) - S_n. \end{aligned}$How big is $S_n$? Well, first we can reckon the $\delta_k$: they are equal to $I_k$ minus the area of the trapezoidal approximation of $I_k$, which is $(\ln(k) + \ln(k+1))/2$:

$\begin{aligned}\delta_k &= \int_k^{k+1}\ln(x)dx - \frac{\ln(k) + \ln(k+1)}{2}\\ &= (k+1)\ln(k+1) - (k+1) - k\ln(k) + k - \frac{\ln(k) + \ln(k+1)}{2}\\ &= -1 + \ln\left(\frac{k+1}{k}\right)(k + 0.5). \end{aligned}$Now you might want to develop the determinant as $\ln(1 + k^{-1}) = k^{-1} + k^{-2}/2 + \dots$: **don't do this**. Instead, do the following dark magic: first, note that $\frac{k+1}{k} = \frac{1 + x}{1-x}$ where $x = (2k+1)^{-1}$. Then, use the analytic formula

so that

$\begin{aligned} \delta_k &= -1 + \frac{1}{x}\left(x + \frac{x^3}{3} + \frac{x^5}{5} + \frac{x^7}{7} + \dotsb \right)\\ &= - \frac{x^2}{3} - \frac{x^4}{5} - \frac{x^6}{7} - \dotsb \\ &= - \frac{1}{3(2k+1)^2} - \frac{1}{5(2k+1)^4} - \frac{1}{7(2k+1)^6} - \dotsb \end{aligned}$We now use this to bound $\delta_k$ below and above.

Since $3,5,7,9\dots$ are all greater than 3,

$\begin{aligned} - \delta_k &< \frac{1}{3(2k+1)^2} + \frac{1}{3(2k+1)^4} + \frac{1}{3(2k+1)^6} + \dotsb \\&= \frac{1}{3(2k+1)^2}\frac{1}{1 - (2k+1)^{-2}}\\&= \frac{1}{12k(k+1)}= \frac{1}{12k}\left(\frac{1}{k} - \frac{1}{k+1} \right). \end{aligned}$On the other hand, $3,5,7,9\dots$ are smaller than $3,3^2, 3^3, 3^4\dots$, so

$\begin{aligned} -\delta_k &> \frac{1}{3(2k+1)^2} + \frac{1}{3^2(2k+1)^4} + \frac{1}{3^3(2k+1)^6} + \dotsb \\&= \frac{1}{3(2k+1)^2}\frac{1}{1 - (3(2k+1))^{-2}}\\ &= \frac{1}{12((k+1/2)^2 - 1/12)}. \end{aligned}$It turns out that $(k+1/2)^2 -1/12 < (k+1/12)(k+1/12 + 1)$ (develop both sides then dismiss some terms), so that

$-\delta_k > \frac{1}{12}\left( \frac{1}{k+1/12} - \frac{1}{k+1/12 + 1} \right).$From the preceding bounds we see that the series $\delta_1 + \delta_2 + \dotsb$ is indeed convergent. If $s$ is its sum, $S_n = s - \sum_{k>n}\delta_k$ and using again the precedings bounds (they are telescoping), we can estimate:

$s - \frac{1}{12n}< S_n < s - \frac{1}{12n+1}$Now go back last line of (1). Take exponentials to get $n!/n^n e^{-n}\sqrt{n} = e^{1-S_n}$. Then, with $c=1-s$ and the estimate above,

$e^{c + \frac{1}{12n+1}}<\frac{n!}{n^n e^{-n}\sqrt{n}}< e^{c + \frac{1}{12n}}.$It's obviously not over, since we didn't get the constant $c$. This, however, is usually done by another means, typically with the Wallis integral asymptotics as they here.

[1] | Herbert Robbins does not have the fame he deserves. He's the Robbins of the Robbins-Munro algorithm, the Lai-Robbins bound on bandit algorithms, the backward algorithm for the secretary problem: three essential contributions to different domains of statistics and computational mathematics. His book What is Mathematics ? with Courant is a masterpiece. |