Stirling's approximation

James Stirling (1692-1770, Scotland)

$\left.\ln N!\right.=\ln 1+\ln 2+\ln 3+...+\ln N=\sum _{k=1}^{N}\ln k$

Because of Euler-MacLaurin formula

$\sum _{k=1}^{N}\ln k=\int _{1}^{N}\ln x\,dx+\sum _{k=1}^{p}{\frac {B_{2k}}{2k(2k-1)}}\left({\frac {1}{n^{2k-1}}}-1\right)+R$

where B₁ = −1/2, B₂ = 1/6, B₃ = 0, B₄ = −1/30, B₅ = 0, B₆ = 1/42, B₇ = 0, B₈ = −1/30, ... are the Bernoulli numbers, and R is an error term which is normally small for suitable values of p.

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle ~\approx \int _{1}^{N}\ln xdx}

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle ~=\left[x\ln x-x\right]_{1}^{N}}

Failed to parse (Conversion error. Server ("https://wikimedia.org/api/rest_") reported: "Cannot get mml. Server problem."): {\displaystyle ~=N\ln N-N+1}

Thus, for large N

$\ln N!\approx N\ln N-N$