= Convergence of the Stirling sequence
{title2=$a_n=\sqrt n(n/e)^n/n!$}
The positive <sequence> $a_n=\sqrt n(n/e)^n/n!$ has logarithmic increments
$$
u_n=\ln\frac{a_{n+1}}{a_n}=\left(n+\frac12\right)\ln\left(1+\frac1n\right)-1.
$$
The <rational lower bound for the natural logarithm> proves $u_n>0$. For $n>1$, the <Taylor series> gives
$$
u_n=\sum_{k=2}^{\infty}\frac{(-1)^k}{n^k}\frac{k-1}{2k(k+1)}.
$$
Since $0<(k-1)/[2k(k+1)]<1/2$, a <geometric series> bounds $u_n<1/[2n(n-1)]\leq n^{-2}$ for $n\geq2$. 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.
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