Question

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

Answer 1

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