Đặt \(A=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}>\dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}>\dfrac{1}{100}+\dfrac{1}{100}+...+\dfrac{1}{100}=\dfrac{25}{100}=\dfrac{1}{4}\)
Do đó: \(A>\dfrac{1}{3}+\dfrac{1}{4}=\dfrac{7}{12}\)(1)
Ta có: \(\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{75}< \dfrac{1}{50}+\dfrac{1}{50}+...+\dfrac{1}{50}=\dfrac{25}{50}=\dfrac{1}{2}\)
\(\dfrac{1}{76}+\dfrac{1}{77}+...+\dfrac{1}{100}< \dfrac{1}{75}+\dfrac{1}{75}+...+\dfrac{1}{75}=\dfrac{25}{75}=\dfrac{1}{3}\)
Do đó: \(A< \dfrac{1}{2}+\dfrac{1}{3}=\dfrac{5}{6}\)(2)
Từ (1) và (2) ta suy ra ĐPCM
