Stirling approximation to a factorial is $$ n! \sim \sqrt{2 \pi n} \left(\frac{n}{e}\right)^n. $$

I wonder what benefit can be got from it?

From computational perspective (I admit I don't know too much about how each arithmetic operation is implemented and which is cheaper than which), a factorial $n!$ contains $n-1$ multiplications. In Stirling's approximation, one also has to compute one division and $n$ multiplications for $\left(\frac{n}{e}\right)^n$, no? Plus two multiplication and one square root for $\sqrt{2 \pi n}$, how does the approximation reduce computation?

There may be considerations from other perspectives. I also would like to know. Please point out your perspective if you can.

Added: For purpose of simplifying analysis by Stirling's approximation, for example, the reply by user1729, my concern is that it is an approximation after all, and even if the approximating expression converges, don't we need to show that the original expression also converges and converges to the same thing as its approximation converges to?

Thanks and regards!

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    $\begingroup$ Well, you can actually calculate $n^{n}$ in $O(\log n)$ multiplications. And if you are interested in $\log(n!)$ instead (as one often is), Stirling's approximation reduces a calculation of $n$ logarithms ($\log n + \log(n-1) + \dots$) to just one ($(n+1/2)\log n - n + O(1)$). $\endgroup$ – mjqxxxx Jan 11 '12 at 14:38
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    $\begingroup$ Approximations can be used in lots of ways. In particular, it makes it easier to compare $n!$ to other functions, to compute limits, etc. And, as others have noted, you don't need to do $n$ multiplications to compute $(\frac{n}{e})^n$. $\endgroup$ – Thomas Andrews Jan 11 '12 at 14:50
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    $\begingroup$ @Tim Computing one logarithm is much more expensive. You wouldn't use Stirling's formula to estimate $6!$ You would use it to estimate $1,000,000!$ In this case, calculating the logarithm is gonna be much faster. $\endgroup$ – Thomas Andrews Jan 11 '12 at 14:51
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    $\begingroup$ For a simple application, consider the number of keys for a substitution cypher, which is 26!. Since $26 \approx 10e, (\frac{26}e)^{26}\approx 10^{26}$ or about 87 bits (as $10^{27}$ would be 90 bits) and to this level we don't care about the square root part. $\endgroup$ – Ross Millikan Jan 11 '12 at 15:18
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    $\begingroup$ Even without using logarithms, you can compute $a^n$ using on the order of $\log n$ multiplications. See: en.wikipedia.org/wiki/Exponentiation_by_squaring $\endgroup$ – Thomas Andrews Jan 11 '12 at 15:29

10 Answers 10


The purpose (as for all asymptotic expressions) is to replace a "complicated" function (in this case the factorial) with some expression which is "simpler". So you might object that $\sqrt{2\pi n} (n/e)^n$ is simpler than $n!$. But if I ask you the question whether $e^n$ or $n!$ grows faster when $n \to \infty$ you might appreciate Stirling's result. Or try to answer the question, how many digits $n!$ has when $n$ is large. Or ...

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    $\begingroup$ Thanks! How do "$n!$ grows faster than $e^n$" and the number of digits of $n!$ make me appreciate Stirngling's result? $\endgroup$ – Tim Jan 11 '12 at 14:51
  • $\begingroup$ I wrote maybe ;-) The only thing I wanted to convey is that for some problems $(n/e)^n$ is a simpler function than $n!$. I do not know how you would decide whether $e^n$ grows faster than $n!$, but $n^n/ e^n \to \infty$ for $n\to \infty$ is rather easy to see. $\endgroup$ – Fabian Jan 11 '12 at 14:57
  • $\begingroup$ My reply has nothing to do with computation but with the fact that you learn something (analytically) about $n!$ when you know Stirling's approximation. Try to answer the question: for which $k\in\mathbb{N}$ the function $\lambda^k/k!$ assumes its maximum when $\lambda \gg 1$ (this is the answer to the question what is the most likely outcome for random variable which Poisson distributed). $\endgroup$ – Fabian Jan 11 '12 at 16:24
  • $\begingroup$ The expression in Stirling's approximation is definitely simpler in the sense that it only involves elementary functions. $\endgroup$ – Mark Jan 11 '12 at 18:00
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    $\begingroup$ @Tim: I wouldn't call it (or its extension - The Gamma function) an elementary (or algebraic) function. $\endgroup$ – Mark Jan 12 '12 at 10:53

One result from Computer Science:

The minimum number of comparisons needed to sort any $n$ items using a comparison-sort is


(this can be seen by taking every possible ordering, and forming binary tree of necessary comparisons of minimum height).

By Stirling's approximation, we have

$$\log_2(n!)$$ $$> \log_2\left(\sqrt{2\pi n}\left(\frac{n}{e}\right)^n\right)$$ $$= n \log_2\left(\frac{n}{e}\right) + \log_2\left(\sqrt{2\pi n}\right)$$

Or, as you will see written it in every Computer Science book ever,

The number of comparisons necessary for any comparison-sort is lower-bounded by $\Omega(n \log_2n)$

Also, a sidenote: OP stated that both $n!$ and $n^n$ require $O(n)$ multiplications to compute. This is false - $n^n$ can be computed quite easily in $O(log_2(n))$


Abraham de Moivre was the person who first introduced Stirling's formula. His friend James Stirling is the one who found that the constant is $\sqrt{2\pi}$; de Moivre only knew it numerically.

Demoivre used it to approximate the probability that the number of heads you get when you toss a coin 1800 times is $x$, for $x$ not too many standard deviations away from 900. He wrote about this in his book titled The Doctrine of Chances (google the title!). The title of the book is in effect 18th-century English for "the theory of probability". The phrase appears again in Thomas Bayes' famous posthumous paper "An essay towards solving a problem in the doctrine of chances" (google that title too). De Moivre derived the bell-shaped curve $$ x\mapsto \text{constant} \cdot e^{-x^2/2} $$ from this formula.


Stirling's approximation is used extensively in Physics, in Boltzmann's description of entropy:

$$ S = k \log_e W$$

Where W is the number of possible permutations of a group of materials, given by

$$W = \frac{N!}{\prod_i N_i!}$$

Taking the the approximation makes the maths much easier.

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    $\begingroup$ As far as I am aware, thermodynamics only need the weaker estimate $\log(n!) \approx n\log n - n$, or even $\log(n!) \approx n\log n$. The $\sqrt{2\pi}$ part of Stirling's formula is irrelevant, and the estimate $\log(n!) \approx n\log n - n$ can be derived in a much simpler way than the methods used to prove Stirling's formula down to the $\sqrt{2\pi}$. $\endgroup$ – KCd Nov 10 '13 at 17:14

I'd like to add several points, which do not seem to have been covered by the existing answers.

  • Stirling's formula reduces the question of computing a "special" function (factorial) to an explicit expression involving only "elementary" functions. This definition of "elementary" may sound pretty arbitrary to you. However, all of the functions occurring in Stirling's formula happen to be implemented in hardware, with guaranteed precision and highly-optimized implementations that you are not likely to beat. Of course, no-one is going to compute $(n/e)^n$ by $n$ multiplications: both $\ln$ and $\exp$ are available as "elementary" functions.

You can argue that direct multiplication may happen to work faster for very small values of $n$. However, the programmer of a general purpose mathematical function would normally prefer a fixed time implementation, if it is at all possible. This has the benefit of being able to estimate run times of computations that rely on computing $n!$. Of course, you can always imagine a situation when $n$ is small "most of the time". A good programmer is then likely to come up with a solution involving look-up table of pre-calculated values for small $n$. Once again, one tends to prefer fixed time implementations.

  • You also seem to be concerned about the accuracy. First, more terms are likely to be needed to achieve required accuracy at moderately small $n$. For example, a common floating point data type has about 15 significant digits; one needs about 5-7 terms in Stirling's expansion to get the necessary accuracy for $n\ge12$. For smaller values of $n$ a look-up table is going to be a fastest solution. So, one would design with a compromise in mind, choosing between allowed storage and worst-case speed. Second, the remainder term, omitted in your question, can actually be used to establish the upper bound on the approximation error, see e.g. book "Special Functions" by Temme (1996). Hence, Sterling's approximation can be (and was) proven to be a sound solution for the problem of computing values of the factorial function.

  • Last, but not least. Stirling's formula happens to work just as well in the case when $n$ is not integer, i.e. for computing the gamma function (usually defined with a trivial shift of the argument). It also works for complex arguments (although nowadays this property is not used very often due to the availability of better approximations). It is, therefore, relevant in a much wider context than just as an approximation for values of the factorial function.


You can use it in Polya's recurrence theorem (I came across it in the first chapter of "Topics in Geometric Group Theory" by de la Harpe). This states,

A simple random walk on $\mathbb{Z}^d$ is recurrent if $d=1$ or $d=2$, and transient if $d\geq 3$.

That is, a walker has probability $1$ of returning to his staring point infinitely often if and only if $d=1$ or $d=2$.

I will give the proof that $d=1$ is recurrent. Look up de la Harpe for the other cases (I'm sure there is a better reference, but you can find these pages of de la Harpe on google books, so it is an available reference...). As you may be able to guess, what comes below is basically just lifted from de la Harpe.

So, there are $2^{2n}$ walks of length $2n$ on $\mathbb{Z}$, and the number of those ending at the origin is $\left( \begin{array}{c} 2n\\ n \end{array} \right)$ (you need precisely $n$ steps to the right and $n$ steps to the left, and you can go either right or left).

This means that the probability, $u_{2n}$, of being at the origin after $2n$ steps is $u_{2n}=\frac{1}{2^{2n}}\left( \begin{array}{c} 2n\\ n \end{array} \right)$. Now, we use Stirling's formula in $\left( \begin{array}{c} 2n\\ n \end{array} \right)$ to get, $$u_{2n}\sim \frac{1}{2^{2n}}\frac{(2n)^{2n}e^{-2n}\sqrt{2\pi 2n}}{n^{2n}e^{-2n}2\pi n}=\frac{1}{\sqrt{\pi n}}$$ (notice how everything just...cancels...).

Noting that $u_{2k+1}=0$ we have that $$\sum^{\infty}_{k=1}u_k=\sum^{\infty}_{n=1}u_{2n}=\frac{1}{\sqrt{\pi}}\sum^{\infty}_{n=1}\frac{1}{\sqrt{n}}=\infty$$ because $\sum^{\infty}_{n=1}\frac{1}{\sqrt{n}}=\infty$, as required.

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    $\begingroup$ This, per se, does not show recurrence, but only that the random number $N$ of visits of $(0,0)$ is not integrable. Recurrence is the stronger property that $N$ is infinite with full probability. $\endgroup$ – Did Jan 11 '12 at 15:16
  • $\begingroup$ Still a typo in the previous to last line. $\endgroup$ – Did Jan 11 '12 at 16:09
  • $\begingroup$ Thanks! My concern about using Stirling's approximation to do analysis is that it is an approximation afterall, and don't you need to show that the original expression also converges and converges to the same thing as it approximation converges to? $\endgroup$ – Tim Jan 11 '12 at 16:57
  • $\begingroup$ Stirling is really used in a comparison to show something converges or diverges; of course the result is extremely accurate with order of one percent error for n = 10 - which is obscenely good. $\endgroup$ – Adam Jan 11 '12 at 21:49
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    $\begingroup$ Typographical error: you say there are $2^{2n}$ walks of length $n$ when you meant walks of length $2n$. $\endgroup$ – KCd Nov 10 '13 at 17:16

As others have noted, Stirling's formula replaces a factorial with a combination of exponents and multiplications. This has the benefit of being easier to work with analytically. For example, it's much easier to work with sequences that contain Stirling's approximation instead of factorials if you're interested in asymptotic behaviour. Taking derivatives of Stirling's formula is fairly easy; factorials, not so much. Also notice that if you're considering asymptotics, then expressions like $$\lim_{n \rightarrow \infty} \frac{a_n}{b_n}$$ where $a_n, b_n$ contain some crazy factorial expressions simplify really nicely.


I don't have a definite answer to your question so I come with one example. Consider the following problem:

What is the probability of obtaining between 45 and 55 heads in 100 tosses of a fair coin?

Clearly, the exact answer is $$P=\frac{1}{2^{100}}\sum_{k=45}^{55}\binom{100}{k}$$ following the binomial distribution, for example. But how large or small is this number? Could we answer if we had only a simple pocket calculator?

First, we note that





hence the computation of $P$ is reduced to the computation of $\dfrac{\binom{100}{50}}{2^{100}}$ (actually, $$P=\dfrac{\binom{100}{50}}{2^{100}}\cdot9.15635\ldots$$ after some multiplications, divisions and additions with the pocket calculator, which took me about 1 minute).

Stirling's formula is used for approximating $\dfrac{\binom{100}{50}}{2^{100}}$.

$$\dfrac{\binom{100}{50}}{2^{100}}=\frac{100!}{2^{100}\cdot\left( 50!\right) ^{2}}\approx\frac{\sqrt{200\pi}\left( \frac{100}{\mathrm{e}}\right) ^{100}}{2^{100}\cdot100\pi\left( \frac{50}{\mathrm{e}}\right) ^{100}}=\frac{1}{\sqrt{50\pi}}\approx0.079788$$ hence $$P\approx\frac{9,15635\ldots}{\sqrt{50\pi}}\approx.7306$$

This is a very good approximation, indeed. The value computed with Mathematica is $.7287\cdots$.


Just mentioning another important application...

Stirling's Formula is an integral part of proving the Prime Number Theorem, specifically used in counting zeros in the critical strip. The Riemann zeta function is modified by multiplying it by a few functions, one of which is the gamma function (specifically, $\Gamma(s/2+1)$); this effectively gets rid of the trivial zeros at the negative even integers. A contour integral of the logarithmic derivative of this modified function then gives the number of zeros inside the contour. Noting that $\Gamma(n+1) = n!$, the piece of the logarithmic derivative coming from the gamma function is approximated by... Stirling.

  • $\begingroup$ You don't need to count zeros of $\zeta(s)$ in the critical strip to prove the Prime Number Theorem. Knowing that there are no zeros on the line ${\rm Re}(s) = 1$ is enough information about the zeros. $\endgroup$ – KCd Feb 14 '14 at 3:59

A Stirling inequality $$(n!)^{\frac{1}{n}} \le \frac{e}{n+1}$$ can be used to derive Carleman's inequality from the AM-GM inequality.


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