Chứng minh rằng

 A = $\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{10}^2}}} < 1$

Chứng minh rằng

 A = $\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{10}^2}}} < 1$

Ta có: $\frac{1}{{{2^2}}} = \frac{1}{{2.2}} < \frac{1}{{1.2}}$

$\frac{1}{{{3^2}}} = \frac{1}{{3.3}} < \frac{1}{{2.3}}$

$\frac{1}{{{4^2}}} = \frac{1}{{4.4}} < \frac{1}{{3.4}}$

…

$\frac{1}{{{{10}^2}}} = \frac{1}{{10.10}} < \frac{1}{{9.10}}$

Nên $\frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ... + \frac{1}{{{{10}^2}}} < \frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{9.10}}$

A <$\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{9.10}}$

Ta lại có: $\frac{1}{{1.2}} + \frac{1}{{2.3}} + \frac{1}{{3.4}} + ... + \frac{1}{{9.10}}$ = 1 - $\frac{1}{2} + \frac{1}{2} - \frac{1}{3} + ... + \frac{1}{9} - \frac{1}{{10}} = 1 - \frac{1}{{10}} = \frac{9}{{10}}$

Vì $\frac{9}{{10}} < 1$ nên A < 1