Question

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

Answer 1

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

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

$[1^2+2^2+3^2...+n^2=\frac{1}{6}n(n+1\left)\right(2n+1\left)\right]$

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

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

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

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

$=\frac{1}{6}\frac{(2+0+0)}{1}=\frac{1}{3}$