Convergence of the Stirling sequence ($a_n=\sqrt{n}(n/e{)}^n/n!$)

The positive sequence $a_n=\sqrt{n}(n/e{)}^n/n!$ has logarithmic incrementsThe rational lower bound for the natural logarithm proves $u_n>0$. For $n>1$, the Taylor series givesSince $0<(k-1)/[2k(k+1)]<1/2$, a geometric series bounds $u_n<1/[2n(n-1)]\le n^{-2}$ for $n\ge 2$. The comparison test for series therefore proves $\sum u_n<\infty$. The finite telescoping sum $\ln a_{n+1}=-1+{\sum }_{k=1}^n{u}_k$ shows that $a_n$ increases to a finite positive limit. This argument establishes the existence of the constant in Stirling's formula; its value $1/\sqrt{2\pi }$ requires an independent identification.