---
tag: 0KC0F51T6A
kind: statement
role: theorem
name: Stirling's formula
aliases: [stirling-approximation]
msc: [41A60, 05A10]
uses: [00SB5EA3VW, 0MVCFMW7S4, 0C1VYVNV2N, 0XCYA4YERS]
sources:
  - url: https://en.wikipedia.org/wiki/Stirling%27s_approximation
    locator: Derivation
same_as:
  wikidata: Q470877
  wikipedia: Stirling's approximation
---

## Statement

As $n \to \infty$,
$$n! \sim \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n.$$
That is, the ratio $\frac{n!}{\sqrt{2 \pi n} \, (n/e)^n}$ tends to $1$.

## Remarks

The relation $\sim$ is [[asymptotic equivalence|00SB5EA3VW]]. The absolute error grows, but the relative error tends to $0$.
Comparing $\ln n! = \sum_{k \le n} \ln k$ with $\int_1^n \ln x \, dx$ gives the form $n! \sim C \sqrt{n} (n/e)^n$. The [[Wallis product|0XCYA4YERS]] gives $C = \sqrt{2\pi}$.
A sharper form is $n! = \sqrt{2 \pi n} (n/e)^n \big(1 + \frac{1}{12 n} + O(n^{-2})\big)$. Mathlib states the limit as `Stirling.tendsto_stirlingSeq_sqrt_pi`.

## History

Abraham de Moivre first stated a less precise form of the result.
In 1730, James Stirling showed that the constant is $\sqrt{2\pi}$.

## Explanations

### L1 human

Stirling's formula gives the size of $n!$ for large $n$.
For example, it gives $10! \approx 3.599 \times 10^6$, while $10! = 3628800$.