阶乘求和的渐近估计与阶乘倒数平方求和:
$n^2\cdot (n-2)!<\sum_{k=1}^{n}k!<(n^2+0.5)\cdot (n-2)!$
$\sum_{k=0}^{\infty}\frac{1}{(k!)^2}=2.27958530233606726743720444081\cdots$