Đặt \(A=\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{100!}\)
Ta thấy:
\(\dfrac{1}{2!}=\dfrac{1}{1.2};\dfrac{1}{3!}=\dfrac{1}{1.2.3}< \dfrac{1}{2.3};...;\dfrac{1}{100!}=\dfrac{1}{1.2...100}< \dfrac{1}{99.100}\)
Cộng vế với vế ta được:
\(A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{99.100}\)
\(\Rightarrow A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A< 1-\dfrac{1}{100}< 1\)
Vậy \(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+...+\dfrac{1}{100!}< 1\) (Đpcm)
\(\dfrac{1}{2!}+\dfrac{1}{3!}+\dfrac{1}{4!}+\dfrac{1}{100!}\)
\(=\left(\dfrac{1}{1!}-\dfrac{1}{2!}\right)+\left(\dfrac{1}{2!}-\dfrac{1}{3!}\right)+\left(\dfrac{1}{3!}-\dfrac{1}{4!}\right)+...+\left(\dfrac{1}{99!}-\dfrac{1}{100!}\right)\)
\(=1-\dfrac{1}{100!}< 1\)
Mn oi!!!!! Giup minh lam 2 bai nay voi!!!!!
