تقریب استرلینگ

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

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

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

${\displaystyle n!=\int _{0}^{\infty }x^{n}e^{-x}\,{\rm {d}}x.}$

${\displaystyle n!=\int _{0}^{\infty }e^{n\ln x-x}\,{\rm {d}}x=e^{n\ln n}n\int _{0}^{\infty }e^{n(\ln y-y)}\,{\rm {d}}y.}$

${\displaystyle \int _{0}^{\infty }e^{n(\ln y-y)}\,{\rm {d}}y\sim {\sqrt {\frac {2\pi }{n}}}e^{-n},}$

${\displaystyle n!=\int _{0}^{\infty }e^{n\ln x-x}\,{\rm {d}}x\sim e^{n\ln n}n{\sqrt {\frac {2\pi }{n}}}e^{-n}.}$

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

${\displaystyle \int _{0}^{\infty }e^{n(\ln y-y)}\,{\rm {d}}y\sim {\sqrt {\frac {2\pi }{n}}}e^{-n}\left(1+{\frac {1}{12n}}\right)}$

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

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

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