Bài 7.
Ta có:
$\frac{1}{2!}=\frac{1}{1.2}$
$\frac{1}{3!}=\frac{1}{2.3}$
$\frac{1}{4!}<\frac{1}{3.4}$
...
$ \frac{1}{100!}<\frac{1}{99.100}$
Suy ra:
$\frac{1}{2!}+\frac{1}{3!}+\frac{1}{4!}+...+\frac{1}{100!}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{99.100}$=$1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}$
=$1-\frac{1}{100}<1$