Chapter 12.5, Problem 50AE

Stirling’s Formula An approximation for $n!$, when $n$ is large, is given by

$n!\approx \sqrt{2n\pi }{\left(\frac{n}{e}\right)}^n\left(1+\frac{1}{12n-1}\right)$

Calculate $12!,20!\text{and}\text{25!}$ on your calculator. Then use Stirling’s formula to approximate $12!,20!\text{and}\text{25!}$