Stirling's approximation
maths
Statement
Stirling’s approximation
For $n \in \mathbb{N}$ we have
$$\sqrt{2\pi n} \left ( \frac{n}{e} \right )^n e^{\frac{1}{12n + 1}} < n! < \sqrt{2\pi n} \left ( \frac{n}{e} \right )^n e^{\frac{1}{12n}}.$$For $n \in \mathbb{N}$ we have
$$\sqrt{2\pi n} \left ( \frac{n}{e} \right )^n e^{\frac{1}{12n + 1}} < n! < \sqrt{2\pi n} \left ( \frac{n}{e} \right )^n e^{\frac{1}{12n}}.$$