\(S=1+\frac{1}{1!}+\frac{1}{2!}+......+\frac{1}{2001!}=1+1+\frac{1}{2!}+\frac{1}{3!}+....+\frac{1}{2001!}< 1+1+\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2000.2001}=2+\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+.....-\frac{1}{2001}< 2+1=3\Rightarrow S< 3\text{(đpcm)}\)
Ta có
\(\dfrac{1}{2!}=\dfrac{1}{1.2};\dfrac{1}{3!}=\dfrac{1}{1.2.3};\dfrac{1}{4!}=\dfrac{1}{1.2.3.4}< \dfrac{1}{3.4};.....;\dfrac{1}{2001!}=\dfrac{1}{1.2.3.4.......2000.2001}< \dfrac{1}{2000.2001}\)
Vậy S<1+1+\(\dfrac{1}{1.2}+\dfrac{1}{2.3}+......+\dfrac{1}{2000.2001}\)
=1+1+1-\(\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+.....+\dfrac{1}{2000}-\dfrac{1}{2001}\)
= 3-\(\dfrac{1}{2001}< 3\)
