Ta có \(\dfrac{1}{1\cdot2}=1-\dfrac{1}{2}\), \(\dfrac{1}{2\cdot3}=\dfrac{1}{2}-\dfrac{1}{3}\),.., \(\dfrac{1}{99\cdot100}=\dfrac{1}{99}-\dfrac{1}{100}\)
\(\Rightarrow A=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{99}-\dfrac{1}{100}=1-\dfrac{1}{100}=\dfrac{99}{100}\)
Xét hiệu \(A-\dfrac{7}{12}=\dfrac{99}{100}-\dfrac{7}{12}=\dfrac{99\cdot3-7\cdot25}{300}=\dfrac{122}{300}>0\) \(\Rightarrow A>\dfrac{7}{12}\)(1)
\(A-\dfrac{5}{6}=\dfrac{99}{100}-\dfrac{5}{6}=\dfrac{99\cdot3-125\cdot5}{300}=\dfrac{-328}{300}< 0\) \(\Rightarrow A< \dfrac{5}{6}\left(2\right)\)
Từ (1) và (2) \(\Rightarrow\dfrac{7}{12}< A< \dfrac{5}{6}\)
Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)
\(=\left(\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}\right)+\left(\dfrac{1}{5\cdot6}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{99\cdot100}\right)\)
\(=\left(\dfrac{1}{2}+\dfrac{1}{12}\right)+\left(\dfrac{1}{5\cdot6}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{99\cdot100}\right)\)
\(=\dfrac{7}{12}+\left(\dfrac{1}{5\cdot6}+\dfrac{1}{7\cdot8}+...+\dfrac{1}{99\cdot100}\right)>\dfrac{7}{12}\)
\(\Leftrightarrow\dfrac{7}{12}< A\)(1)
Ta có: \(A=\dfrac{1}{1\cdot2}+\dfrac{1}{3\cdot4}+\dfrac{1}{5\cdot6}+...+\dfrac{1}{99\cdot100}\)
\(=\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+\dfrac{1}{5}-\dfrac{1}{6}+...+\dfrac{1}{99}-\dfrac{1}{100}\)
\(=\left(1+\dfrac{1}{3}+\dfrac{1}{5}+...+\dfrac{1}{99}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+\dfrac{1}{4}+\dfrac{1}{5}+\dfrac{1}{6}+...+\dfrac{1}{99}+\dfrac{1}{100}\right)-2\cdot\left(\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{6}+...+\dfrac{1}{100}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{100}\right)-\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{50}\right)\)
\(=\dfrac{1}{51}+\dfrac{1}{52}+...+\dfrac{1}{100}\)
\(=\left(\dfrac{1}{51}+\dfrac{1}{52}+\dfrac{1}{53}+...+\dfrac{1}{60}\right)+\left(\dfrac{1}{61}+\dfrac{1}{62}+\dfrac{1}{63}+...+\dfrac{1}{70}\right)+...+\left(\dfrac{1}{91}+\dfrac{1}{92}+\dfrac{1}{93}+...+\dfrac{1}{100}\right)\)
\(\Leftrightarrow A< \dfrac{1}{50}\cdot10+\dfrac{1}{60}\cdot10+\dfrac{1}{70}\cdot10+\dfrac{1}{80}\cdot10+\dfrac{1}{90}\cdot10\)
\(\Leftrightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}+\dfrac{1}{8}+\dfrac{1}{9}\)
\(\Leftrightarrow A< \dfrac{1}{5}+\dfrac{1}{6}+\dfrac{1}{7}\cdot3\)
\(\Leftrightarrow A< \dfrac{167}{210}< \dfrac{175}{210}=\dfrac{5}{6}\)
hay \(A< \dfrac{5}{6}\)(2)
Từ (1) và (2) suy ra \(\dfrac{7}{12}< A< \dfrac{5}{6}\)(đpcm)
