Chứng tỏ rằng :

- Hướng dẫn giải

$a)\frac{1}{3.4}+\frac{1}{4.5}+\mathrm{...}+\frac{1}{19.20}=\frac{1}{3}-\frac{1}{20}=\frac{17}{60}<\frac{1}{2}$

$b)\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+\mathrm{...}+\frac{3}{97.100}=1-\frac{1}{100}<1$

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

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