Stirling's approximation

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James Stirling (1692-1770, Scotland)

lnN!=ln1+ln2+ln3+...+lnN=∑k=1Nlnk

Because of Euler-MacLaurin formula

∑k=1Nlnk=∫1Nlnxdx+∑k=1pB2k2k(2k−1)(1n2k−1−1)+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.

Then, for large N,

lnN!≈∫1Nlnxdx=NlnN−N