Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

This is a naive question about history.

My understanding is that Stirling's formula or something trivially equivalent to it first appeared in an early edition of Abraham de Moivre's book The Doctrine of Chances (and maybe in a journal article he wrote before that?), provided that we understand "Stirling's formula" to mean that $$ \lim_{n\to\infty} \frac{\sqrt{n}\cdot n^n e^{-n}}{n!} = \text{some finite non-zero number}. $$ If I'm not mistaken, de Moivre computed this number numerically and it was James Stirling who later showed just which number it is, and then de Moivre included it in a later edition of his book.

It's been a while since I read about all this and I don't remember enough detail to be 100% sure I've got all of the above right.

But today the term "Stirling's formula" seems to be used to refer to an asymptotic expansion.

How much of that series called "Stirling's formula" can be attributed to Stirling? And who should get credit for what he shouldn't get credit for?

share|cite|improve this question
The phrase "converges in a sense other than the one that would conflict with calling it divergent" is confusing. Can you write that part more clearly? And as for what seems to be a second meaning of the term Stirling's formula, what is your source? In Whittaker and Watson's Modern Analysis, the asymptotic expansion for the log of the Gamma function is called Stirling's series, not Stirling's formula. – KCd Feb 14 '12 at 16:23
The Stirling series is an asymptotic expansion, not a convergent series for an fixed value of $n$, as you can read here. This means taking more terms gives better and better asymptotic estimates as $n\to\infty$. I don't think the series is actually due to Stirling; no surprise there. – Marc van Leeuwen Feb 14 '12 at 18:12
There seems to be general agreement that, as you write, the evaluation of the constant as $\sqrt{2\pi}$ is due to Stirling. More interestingly perhaps, the evaluation of $\int_0^\infty e^{-x^2}dx$ is due to Stirling. In each case, the main tool was the Wallis product. – André Nicolas Feb 14 '12 at 18:42
OK, I've rephrased it to say something simpler. – Michael Hardy Feb 14 '12 at 18:55
@MichaelHardy: Should the last few words read: "for what he isn't getting credit for?" – Aryabhata Feb 14 '12 at 19:58
up vote 2 down vote accepted

Stirling's original result is the expansion $$ \log n! \sim \left( {n + \frac{1}{2}} \right)\log \left( {n + \frac{1}{2}} \right) - \left( {n + \frac{1}{2}} \right) + \frac{1}{2}\log \left( {2\pi } \right) - \frac{1}{{24\left( {n + \frac{1}{2}} \right)}} + \cdots , $$ as $n\to \infty$. De Moivre's version is that $$ \log n! \sim \left( {n + \frac{1}{2}} \right)\log n - n + \frac{1}{2}\log \left( {2\pi } \right) + \frac{1}{{12n}} - \cdots , $$ for large $n$. A nice reference for the topic is

A. Hald, A History of Probability and Statistics and Their Applications Before 1750. John Wiley $\&$ Sons, New York, 1990.

share|cite|improve this answer
Does $a_n\sim b_n$ in this context mean the difference approaches $0$ as $n\to\infty$? – Michael Hardy Aug 6 '13 at 14:16
No it does not. It means a stronger fact. We can write De Moivre's result as $$ \log \left( {\frac{{n!}}{{n^n e^{ - n} \sqrt {2\pi n} }}} \right) \sim \frac{1}{{12n}} - \frac{1}{{360n^3 }} + \frac{1}{{1260n^5 }} - \cdots \; . $$ The $\sim$ means that the RHS is an asymptotic expansion of the LHS. Here is the definition: – Gary Aug 6 '13 at 15:08
Did de Moivre actually know the value of the constant term when he wrote his first paper on this? I think some version of that paper later became a chapter of his book, but he I think he knew more by that time. – Michael Hardy Aug 6 '13 at 19:19
Stirling told him the value of the constant, but after some time, De Moivre figured it out himself using Wallis' product. – Gary Aug 6 '13 at 19:35

As you note (Wikipedia)

The formula was first discovered by Abraham de Moivre in the form

$$n!\sim \text{constant}\cdot n^{n+1/2} e^{-n}.$$

De Moivre gave an expression for the constant in terms of its natural logarithm. Stirling's contribution consisted of showing that the constant is $\sqrt{2\pi}$. The more precise versions are due to Jacques Binet.

When it comes to getting credit, you should check that lemma that says that when a finding is attributted to someone, it isn't usually the real discoverer (ironically enough this finding was proved to be discovered by yet anoher person before, so the finding itself is an example of this controversy).

I guess the credit is given to him for his asymptotic expansion in series, which gives the "little" approximation:

$$n! \sim {n^n}{e^{ - n}}\sqrt {2\pi n} $$

As to how the approximation is found, I answered some days ago the following:

If you're familiar with [Wallis infinite product]

$$\mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!!}}{{\left( {2n - 1} \right)!!}}\frac{1}{{\sqrt n }} = \sqrt \pi $$

Then you can use

$$\eqalign{ & \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!{!^2}}}{{\left( {2n} \right)!}}\frac{1}{{\sqrt n }} = \sqrt \pi \cr & \mathop {\lim }\limits_{n \to \infty } \frac{{{2^{2n}}{{\left( {n!} \right)}^2}}}{{\left( {2n} \right)!}}\frac{1}{{\sqrt n }} = \sqrt \pi \cr} $$

Now you can check that

$$\alpha = \mathop {\lim }\limits_{n \to \infty } \frac{{n!{e^n}}}{{{n^n}\sqrt n }} = \mathop {\lim }\limits_{n \to \infty } \frac{{\left( {2n} \right)!{e^{2n}}}}{{{{\left( {2n} \right)}^{2n}}\sqrt {2n} }}$$

exists. Then square the first expression and divide by the latter to get

$$\alpha = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n!} \right)}^2}{e^{2n}}}}{{{n^{2n}}n}}\frac{{{{\left( {2n} \right)}^{2n}}\sqrt {2n} }}{{\left( {2n} \right)!{e^{2n}}}} = \mathop {\lim }\limits_{n \to \infty } \frac{{{{\left( {n!} \right)}^2}{2^{2n}}\sqrt 2 }}{{\left( {2n} \right)!\sqrt n }} = \sqrt {2\pi } $$

Thus you have that

$$\mathop {\lim }\limits_{n \to \infty } \frac{{n!{e^n}}}{{{n^n}\sqrt {2n} }} = \sqrt \pi $$


$$n! \sim {n^n}{e^{ - n}}\sqrt {2\pi n} $$

share|cite|improve this answer
"that lemma that says that when a finding is attributed to someone, it isn't usually the real discoverer" - that would be "Stigler's law". – J. M. May 29 '13 at 3:22
@J.M. Or "Arnold's principle" :) – Artem May 29 '13 at 3:33
@Artem, thank you, and now we have amply demonstrated Peter's mention of "ironically enough, this finding was proven to be discovered by yet another person before". ;) – J. M. May 29 '13 at 3:34
@J.M.: Already, Stigler's law says Stigler attributed it to Merton. BTW, a related question is this one. – ShreevatsaR Aug 6 '13 at 11:48

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.