\(\frac{1}{2\cdot2}< \frac{1}{1\cdot2};...;\frac{1}{9\cdot9}< \frac{1}{8\cdot9}\)
\(\Rightarrow S=\frac{1}{2^2}+...+\frac{1}{9^2}< \frac{1}{1\cdot2}+...+\frac{1}{8\cdot9}=1-\frac{1}{2}+...+\frac{1}{8}-\frac{1}{9}=1-\frac{1}{9}=\frac{8}{9}\left(1\right)\)
\(\frac{1}{2\cdot2}>\frac{1}{2\cdot3};...;\frac{1}{9\cdot9}>\frac{1}{9\cdot10}\)
\(\Rightarrow S=\frac{1}{2^2}+...+\frac{1}{9^2}>\frac{1}{2\cdot3}+...+\frac{1}{9\cdot10}=\frac{1}{2}-\frac{1}{3}+...+\frac{1}{9}-\frac{1}{10}=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\left(2\right)\)
Từ (1)(2) => đpcm