$\frac{1}{2!} +\frac{2}{3!} +\frac{3}{4!} +... + \frac{99}{100!}
= (\frac{1}{1!} -\frac{1}{2!}) + (\frac{1}{2!} - \frac{1}{3!}) + (\frac{1}{3!} -\frac{1}{4!}) + ... + (\frac{1}{99!} -\frac{1}{100!})
=1 - \frac{1}{100!} <1$
Vậy 2!1+3!2+4!3+...+100!99<1
2!1+3!2+4!3+⋯+100!99=2!2−1+3!3−1+4!4−1+⋯+100!100−1=1!1−2!1+2!1−3!1+3!1−4!1+⋯+99!1−100!1=1−100!1<1