Ta có
$1^2 + 2^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$
Do đó
$\lim (1^2 + 2^2 + \cdots + n^2) = \lim \left[ \dfrac{n(n+1)(2n+1)}{6} \right]$
$= \lim n^3\left[ \dfrac{\left( 1 + \frac{1}{n} \right) \left( 2 + \frac{1}{n} \right)}{6} \right]$
$= +\infty . \dfrac{2}{6} = +\infty$.