Ta có
$1^2 + 2^2 + \cdots + n^2 = \dfrac{(n^2+n)(2n+1)}{6} = \dfrac{2n^3+3n^2 +n}{6}$
Vậy ta có
$\lim \sqrt{\dfrac{1^2 + 2^2 + \cdots + n^2}{(n^2+n)(n+2)}} = \lim \sqrt{\dfrac{(n^2 + n)(2n+1)}{(n^2 + n)(n+2)}} = \lim \sqrt{\dfrac{2n+1}{n+2}} = \sqrt{2}$