Question

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

• A. 1
• B. $\frac{1}{2}$
• C. $\frac{1}{3}$
• D. 0

Answer 1

The correct option is C

$\frac{1}{3}$

${lim}_{n\to \infty }\frac{1^2+2^2+3^2\cdots n^2}{n^3}$

$={lim}_{n\to \infty }\frac{\sum n^2}{n^3}$
$={lim}_{n\to \infty }\frac{n(n+1)(2n+1)}{6n^3}$
$={lim}_{n\infty \infty }\frac{(n+1)(2n+1)}{6n^2}$
Dividing the numerato5r and the denominator by $n^2,$ we get :
$={lim}_{n\to \infty }\frac{\frac{(n+1)}{n}\times \frac{(2n+1)}{n}}{6}$
$={lim}_{n\to \infty }\frac{(1+\frac{1}{n})\times (2+\frac{1}{n})}{6}$
$\Rightarrow \frac{2}{6}=\frac{1}{3}$