Ta có
$A = \dfrac{1}{3} + \dfrac{1}{3^2} + \dfrac{1}{4}^2 + \cdots + \dfrac{1}{2009^2}$
$< \dfrac{1}{3} + \dfrac{1}{2.3} + \dfrac{1}{3.4} + \cdots + \dfrac{1}{2008.2009}$
$= \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{3} + \dfrac{1}{3} - \dfrac{1}{4} + \cdots + \dfrac{1}{2008} - \dfrac{1}{2009}$
$= \dfrac{1}{3} + \dfrac{1}{2} - \dfrac{1}{2009}$
$= \dfrac{5}{6} - \dfrac{1}{2009} < 1$
Vậy $A<1$.