proof of Stirling’s approximation

Computing the Taylor expansion with remainder of the functions $\log$ and $x\mapsto x\log x-x$, we have

$(n+1)\log (n+1)-\log (n+1)$ $=$ $n\log n-n+\log n+{\displaystyle \frac{1}{2n}}+{\displaystyle \frac{1}{6{\xi }_n^2}}$

$\log (n+1)$ $=$ $\log n+{\displaystyle \frac{1}{n}}-{\displaystyle \frac{1}{2{\eta }_n^2}}$

where $n\le {\xi }_n\le n+1$ and $n\le {\eta }_n\le n+1$. Summing the first equation from $1$ to $n-1$, we have

$$n\log n-n=-1+\log (n-1)!+\frac{1}{2}\sum _{m=1}^{n-1}\frac{1}{m}+\frac{1}{6}\sum _{m=1}^{n-1}\frac{1}{{\xi }_n^2}.$$

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