Đặt A=\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\)

Vì \(\dfrac{1}{101}\)>\(\dfrac{1}{102}\)>\(\dfrac{1}{103}\)>...>\(\dfrac{1}{300}\)

=>(\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{200}\))+(\(\dfrac{1}{201}\)+\(\dfrac{1}{202}\)+\(\dfrac{1}{203}\)+...+\(\dfrac{1}{300}\)) > (\(\dfrac{1}{200}\)+\(\dfrac{1}{200}\)+\(\dfrac{1}{200}\)+...+\(\dfrac{1}{200}\))+(\(\dfrac{1}{300}\)+\(\dfrac{1}{300}\)+\(\dfrac{1}{300}\)+...+\(\dfrac{1}{300}\)) =>\(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) > \(\dfrac{1}{200}\).100 +\(\dfrac{1}{300}\) .100

=> A > \(\dfrac{1}{2}+\dfrac{1}{3}\)

=> A > \(\dfrac{5}{6}\) Mà \(\dfrac{5}{6}\)>\(\dfrac{2}{3}\)=> A > \(\dfrac{2}{3}\) Vậy \(\dfrac{1}{101}\)+\(\dfrac{1}{102}\)+\(\dfrac{1}{103}\)+...+\(\dfrac{1}{300}\) >\(\dfrac{2}{3}\)