proof of Stirling’s approximation


Computing the Taylor expansionMathworldPlanetmath with remainder of the functions log and xxlogx-x, we have

(n+1)log(n+1)-log(n+1) = nlogn-n+logn+12n+16ξn2
log(n+1) = logn+1n-12ηn2

where nξnn+1 and nηnn+1. Summing the first equation from 1 to n-1, we have

nlogn-n=-1+log(n-1)!+12m=1n-11m+16m=1n-11ξn2.
Title proof of Stirling’s approximation
Canonical name ProofOfStirlingsApproximation
Date of creation 2014-05-08 22:09:30
Last modified on 2014-05-08 22:09:30
Owner rspuzio (6075)
Last modified by rspuzio (6075)
Numerical id 6
Author rspuzio (6075)
Entry type Proof
Classification msc 68Q25
Classification msc 30E15