Đáp án:
$\begin{array}{l}
A = \dfrac{1}{{{2^2}}} + \dfrac{1}{{{3^2}}} + \dfrac{1}{{{4^2}}} + ... + \dfrac{1}{{{n^2}}}\\
Do:1.2 < {2^2}\\
\Rightarrow \dfrac{1}{{{2^2}}} < \dfrac{1}{{1.2}}\\
TT:\dfrac{1}{{{3^2}}} < \dfrac{1}{{2.3}};\dfrac{1}{{{4^2}}} < \dfrac{1}{{3.4}};...;\dfrac{1}{{{n^2}}} < \dfrac{1}{{\left( {n - 1} \right).n}}\\
\Rightarrow A < \dfrac{1}{{1.2}} + \dfrac{1}{{2.3}} + \dfrac{1}{{3.4}} + ... + \dfrac{1}{{\left( {n - 1} \right).n}}\\
\Rightarrow A < 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + .. + \dfrac{1}{{n - 1}} - \dfrac{1}{n}\\
\Rightarrow A < 1 - \dfrac{1}{n} < 1\\
Vậy\,A < 1
\end{array}$
