Chứng tỏ rằng: a) 1/1.2 + 1/2.3 + ... + 1/9.10 < 1 b) 3/2.5 + 3/5.8

* Hướng dẫn giải

$a)\frac{1}{1.2}+\frac{1}{2.3}+\mathrm{...}+\frac{1}{9.10}=\frac{9}{10}<1.$

$b)\frac{3}{2.5}+\frac{3}{5.8}+\frac{3}{8.11}+\frac{3}{11.14}+\frac{3}{14.17}+\frac{3}{17.20}=\frac{1}{2}-\frac{1}{20}<\frac{1}{2}.$

$c)Tacó \frac{1}{2^2} <\frac{1}{1.2};\frac{1}{3^2}<\frac{1}{2.3};\frac{1}{4^2}<\frac{1}{3.4};\mathrm{...};\frac{1}{{25}^2}<\frac{1}{24.25}.$

$Dođó,\frac{1}{2^2} +\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{{25}^2}<\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\mathrm{...}+\frac{1}{24.25}<1.$