Question

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

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

Answer 1

The correct option is D $\frac{1}{3}$

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

$=\underset{n\to \infty }{lim}\frac{\frac{n(n+1)(2n+1)}{6}}{n^3}$
$=\underset{n\to \infty }{lim}\frac{(1+\frac{1}{n})(2+\frac{1}{n})}{6}$

$=\frac{2}{6}$

$=\frac{1}{3}$