Có: \(\dfrac{1}{101}>\dfrac{1}{200};\dfrac{1}{102}>\dfrac{1}{200};...;\dfrac{1}{199}>\dfrac{1}{200};\dfrac{1}{200}=\dfrac{1}{200}\)
Số các phân số trong \(\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}\) là
`(200-101):1+1=100` (số)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}>\dfrac{1}{200}.100\)
\(\Rightarrow\dfrac{1}{101}+\dfrac{1}{102}+\dfrac{1}{103}+...+\dfrac{1}{200}>\dfrac{1}{2}\left(đpcm\right)\)
đặt phân số dạng chung : \(\dfrac{1}{k+1}\) ( \(101\le k\le200\) ) .
Có : \(\dfrac{1}{k+1}\) > \(\dfrac{1}{k\left(k+1\right)}\) = \(\dfrac{1}{k}-\dfrac{1}{k+1}\)
biểu thức trên > \(\dfrac{1}{101}-\dfrac{1}{102}+\dfrac{1}{102}-\dfrac{1}{103}+...+\dfrac{1}{200}-\dfrac{1}{201}\)
= \(\dfrac{1}{101}-\dfrac{1}{201}\) > \(\dfrac{1}{2}\)
