# weaker version of Stirling’s approximation

One can prove a weaker version of Stirling’s approximation without appealing to the gamma function. Consider the graph of $\ln x$ and note that

 $\ln(n-1)!\leq\int_{1}^{n}\ln x\,\mathrm{d}x\leq\ln n!$

But $\int\ln x\,\mathrm{d}x=x\ln x-x$, so

 $\ln(n-1)!\leq n\ln n-n+1\leq\ln n!$

and thus

 $n\ln n-n+1+\ln n\geq\ln(n-1)!+\ln n=\ln n!\geq n\ln n-n+1$

so

 $\ln n-1+\frac{1}{n}+\frac{\ln n}{n}\geq\frac{1}{n}\ln n!\geq\ln n-1+\frac{1}{n}$

As $n$ gets large, the expressions on either end approach $\ln n-1$, so we have

 $\frac{1}{n}\ln n!\approx\ln n-1$

Multiplying through by $n$ and exponentiating, we get

 $n!\approx n^{n}e^{-n}$
Title weaker version of Stirling’s approximation WeakerVersionOfStirlingsApproximation 2013-03-22 16:25:21 2013-03-22 16:25:21 rm50 (10146) rm50 (10146) 7 rm50 (10146) Result msc 41A60 msc 30E15 msc 68Q25