تقریب استرلینگ

${\displaystyle n!\approx A(n)={\left(\frac{n}{e}\right)}^n\sqrt{2\pi n}}$

${\displaystyle E=\frac{|n!-A(n)|}{n!}}$

${\displaystyle \frac{1}{12n-1}}$

${\displaystyle n!={\int }_0^{\mathrm{\infty }}{x}^n{e}^{-x}\mathrm{d}x.}$

${\displaystyle n!={\int }_0^{\mathrm{\infty }}{e}^{n\ln x-x}\mathrm{d}x=e^{n\ln n}n{\int }_0^{\mathrm{\infty }}{e}^{n(\ln y-y)}\mathrm{d}y.}$

${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{n(\ln y-y)}\mathrm{d}y\sim \sqrt{\frac{2\pi }{n}}{e}^{-n},}$

${\displaystyle n!={\int }_0^{\mathrm{\infty }}{e}^{n\ln x-x}\mathrm{d}x\sim e^{n\ln n}n\sqrt{\frac{2\pi }{n}}{e}^{-n}.}$

${\displaystyle n!\sim \sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n.}$

${\displaystyle {\int }_0^{\mathrm{\infty }}{e}^{n(\ln y-y)}\mathrm{d}y\sim \sqrt{\frac{2\pi }{n}}{e}^{-n}\left(1+\frac{1}{12n}\right)}$

${\displaystyle n!\sim e^{n\ln n}n\sqrt{\frac{2\pi }{n}}{e}^{-n}\left(1+\frac{1}{12n}\right)=\sqrt{2\pi n}{\left(\frac{n}{e}\right)}^n\left(1+\frac{1}{12n}\right).}$

${\displaystyle ln(A(15))=15(ln(15)-1)+\frac{1}{2}(ln(2\pi )+ln(15))\approx 27.89371}$

${\displaystyle 15!\approx A(15)=e^{ln(A(15))}\approx 1300420000000}$