James Stirling (1692-1770, Scotland)

Because of Euler-MacLaurin formula

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.

![{\displaystyle ~=\left[x\ln x-x\right]_{1}^{N}}](https://wikimedia.org/api/rest_v1/media/math/render/svg/2ba6f8a16f83420e34962f787fd0668a13baa710)

Thus, for large N
