Question

$\underset{n\to \infty }{\lim }\left[\frac{1^2+2^2+...+n^2}{n^3}\right]$

Answer 1

$$\begin{gathered} =\underset{n\to \infty }{\lim }\left[\frac{1^2+2^2+...+n^2}{n^3}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{n\left(n+1\right)\left(2n+1\right)}{6n^3}\right] \\ =\underset{n\to \infty }{\lim }\left[\frac{\left(n+1\right)\left(2n+1\right)}{6n^2}\right] \\ =\underset{n\to \infty }{\lim }\left[\left(\frac{n+1}{n}\right)\left(\frac{2n+1}{n}\right)\times \frac{1}{6}\right] \\ =\underset{n\to \infty }{\lim }\left[\left(1+\frac{1}{n}\right)\left(2+\frac{1}{n}\right)\times \frac{1}{6}\right] \\ =\frac{2}{6} \\ =\frac{1}{3} \end{gathered}$$