Question

$\underset{n\to \infty }{\lim }\frac{1^2+2^2+3^2+...+n^2}{n^3}$ is equal to

(a) 1
(b) 1/2
(c) 1/3
(d) 0

Answer 1

(c) 1/3

$$\begin{gathered} \underset{n\to \infty }{\lim }\frac{1^2+2^2+3^2.....n^2}{n^3} \\ =\underset{n\to \infty }{\lim }\frac{\Sigma n^2}{n^3} \\ =\underset{n\to \infty }{\lim }\frac{n\left(n+1\right)\left(2n+1\right)}{6n^3} \\ =\underset{n\to 0}{\lim }\frac{\left(n+1\right)\left(2n+1\right)}{6n^2} \\ {\text{Dividing the numerator and the denominator by n}}^2,\text{we get:} \\ \underset{n\to \infty }{\lim }\frac{{\displaystyle \frac{\left(n+1\right)}{n}}\times {\displaystyle \frac{\left(2n+1\right)}{n}}}{6} \\ =\underset{n\to \infty }{\lim }\frac{\left(1+{\displaystyle \frac{1}{n}}\right)\left(2+{\displaystyle \frac{1}{n}}\right)}{6} \\ \Rightarrow \frac{2}{6}=\frac{1}{3} \end{gathered}$$