Ta có
$A = \dfrac{1}{2^2} + \dfrac{1}{2^3} + \cdots + \dfrac{1}{n^2}$
$< \dfrac{1}{1.2} + \dfrac{2.3} + \cdots + \dfrac{1}{(n-1)n}$
$= 1 - \dfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \cdots + \dfrac{1}{n-1} - \dfrac{1}{n}$
$= 1 - \dfrac{1}{n} < 1$
Vậy $A < 1$.