Ta có
$\dfrac{1}{2^2} < \dfrac{1}{1.2}$
$\dfrac{1}{3^2} < \dfrac{1}{2.3}$
...
$\dfrac{1}{2009^2} < \dfrac{1}{2008.2009}$
Vậy ta có
$\dfrac{1}{2^2} + \cdots + \dfrac{1}{2009^2} < \dfrac{1}{1.2} + \dfrac{2.3} + \cdots + dfrac{1}{2008.2009}$
$< 1 - \đfrac{1}{2} + \dfrac{1}{2} - \dfrac{1}{3} + \cdots + \dfrac{1}{2008} - \dfrac{1}{2009}$
$= 1 - \dfrac{1}{2009} < 1$