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 (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ~\approx \int_1^N \ln x dx}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ~= \left[ x \ln x - x \right]_1^N}

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle ~= N \ln N -N +1}

Thus, for large N

Failed to parse (SVG (MathML can be enabled via browser plugin): Invalid response ("Math extension cannot connect to Restbase.") from server "https://wikimedia.org/api/rest_v1/":): {\displaystyle \ln N! \approx N \ln N -N}