Chứng tỏ rằng : 

- Hướng dẫn giải

$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\mathrm{...}+\frac{1}{9^2}>\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\mathrm{...}+\frac{1}{9.10}=\frac{2}{5}$

$\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\mathrm{...}+\frac{1}{9^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{8.9}=\frac{8}{9}$