User blog comment:ArtismScrub/Defining a constant, maybe?/@comment-11227630-20171218153216

χ = 1.

Stirling's approximation results $$n!\approx(\frac{n}{e})^n\sqrt{2\pi n}$$, then $$c_n\approx\frac{n\ln n-n+\frac{1}{2}\ln(2\pi n)}{\ln n}$$, so $$\frac{n}{c_n}\approx\frac{n\ln n}{n\ln n-n+\frac{1}{2}\ln(2\pi n)}$$. And $$\lim_{n\rightarrow\infty}\frac{n}{c_n}=\lim_{n\rightarrow\infty}\frac{1+\ln n}{\ln n+\frac{1}{2n}}=1$$.