Trước hết ta biến đổi A thành \(\frac{1}{51}+\frac{1}{52}+\frac{1}{53}+...+\frac{1}{100}\)
Do đó : \(A=\left[\frac{1}{51}+\frac{1}{52}+...+\frac{1}{75}\right]+\left[\frac{1}{76}+\frac{1}{77}+...+\frac{1}{100}\right]\)
Ta có : \(\frac{1}{51}>\frac{1}{52}>...>\frac{1}{75},\frac{1}{76}>\frac{1}{77}>...>\frac{1}{100}\)nên
\(A>\frac{1}{75}\cdot25+\frac{1}{100}\cdot25=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)
\(A< \frac{1}{51}\cdot25+\frac{1}{76}\cdot25< \frac{1}{50}\cdot25+\frac{1}{75}\cdot25=\frac{1}{2}+\frac{1}{3}=\frac{5}{6}\)
Vậy \(\frac{7}{12}< A< \frac{5}{6}\)
Cách biến đổi :
Đặt \(A=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
Dễ thấy \(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}-2\left[\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{100}\right]\)
\(=\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{99}+\frac{1}{100}-\left[\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{50}\right]\)
\(=\frac{1}{51}+\frac{1}{52}+...+\frac{1}{100}\)
