Ta có: A= \(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{90}\)
\(A=\left(\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}\right)\)
A= B+C
Ta có: \(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>\dfrac{1}{60}+\dfrac{1}{60}+...+\dfrac{1}{60}\)
\(B=\dfrac{1}{31}+\dfrac{1}{32}+...+\dfrac{1}{60}>30.\dfrac{1}{60}=\dfrac{1}{2}\) (1)
Lại có: \(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>\dfrac{1}{90}+\dfrac{1}{90}+...+\dfrac{1}{90}\)
\(C=\dfrac{1}{61}+\dfrac{1}{62}+...+\dfrac{1}{90}>30.\dfrac{1}{90}=\dfrac{1}{3}\) (2)
Từ (1) và (2) => \(A>\dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)
Vậy \(A>\dfrac{5}{6}\)
