Question

chứng tỏ\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{17}< 2\)

Answer 1

Đặt A=\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{17}\)\(=\left(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}\right)+\left(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{17}\right)\)

Có: \(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{10}< \frac{1}{5}.6=\frac{6}{5}\)(1)

     \(\frac{1}{11}+\frac{1}{12}+...+\frac{1}{17}< \frac{1}{11}.7=\frac{7}{11}\)(2)

Từ (1) và (2) suy ra: A\(< \frac{6}{5}+\frac{7}{11}=\frac{101}{55}\)

Lại có: \(\frac{101}{55}< \frac{110}{55}=2\)

Suy ra: A<2 (đpcm)