# proof of Stirling’s approximation

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

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

where $n\leq\xi_{n}\leq n+1$ and $n\leq\eta_{n}\leq n+1$. 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 ProofOfStirlingsApproximation 2014-05-08 22:09:30 2014-05-08 22:09:30 rspuzio (6075) rspuzio (6075) 6 rspuzio (6075) Proof msc 68Q25 msc 30E15