Ta có
$\underset{n \to +\infty}{\lim} \dfrac{2.1^2 + 3.2^2 + \cdots + (n+1)n^2}{n^4} = \underset{n \to +\infty}{\lim} \dfrac{(1 + 1).1^2 + (2+1)2^2 + \cdots + n^3 + n^2}{n^4}$
$= \underset{n \to +\infty}{\lim} \dfrac{(1^2 + 2^2 + \cdots + n^2) + (1^3 + 2^3 + \cdots + n^3)}{n^4}$
Lại có
$1^2 + \cdots + n^2 = \dfrac{n(n+1)(2n+1)}{6}$
và
$1^3 + \cdots + n^3 = \dfrac{n^2(n+1)^2}{4}$
Vậy ta có tử số bằng
$\dfrac{n(n+1)(2n+1)}{6} + \dfrac{n^2(n+1)^2}{4} = \dfrac{2n(n+1)(2n+1) + 3n^2(n+1)^2}{12}$
Do đó
$\dfrac{(1^2 + 2^2 + \cdots + n^2) + (1^3 + 2^3 + \cdots + n^3)}{n^4} = \dfrac{2n(n+1)(2n+1) + 3n^2(n+1)^2}{12n^4}$
$= \dfrac{ \frac{2}{n} (1 + \frac{1}{n}) (2 + \frac{1}{n}) + 3(1 + \frac{1}{n})^2}{12}$
Do đó
$\underset{n \to +\infty}{\lim} \dfrac{2.1^2 + 3.2^2 + \cdots + (n+1)n^2}{n^4} = \underset{n \to +\infty}{\lim} \dfrac{ \frac{2}{n} (1 + \frac{1}{n}) (2 + \frac{1}{n}) + 3(1 + \frac{1}{n})^2}{12}$
$= \dfrac{3}{12} = \dfrac{1}{4}$
