Question

Evaluate the limit to infinity
$\underset{x\to \infty }{lim}\frac{1^2+2^2+----+n^2}{n^3}=\frac{1}{3}$

Answer 1

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

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

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

$=\frac{1}{3}$