Finding Limits 4

$\large\lim_{n\to\infty}\frac{1}{n}(\frac{1^2}{n^2}+\frac{2^2}{n^2}+\frac{3^2}{n^2}+...+\frac{n^2}{n^2})=$

$\large\lim_{n\to\infty}\frac{1^2+2^2+3^2+...+n^2}{n^3}=$

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

$\large\lim_{n\to\infty}\frac{(1+1/n)(2+1/n)}{6}=\boxed{\frac{1}{3}}= \int_0^1 x^2dx$

1.- $\displaystyle \lim_{n \to \infty} \frac{1}{n} \cdot \sum_{m = 1}^n \frac{m^2}{n^2} = \int_0^1 x^2 dx = \left[\frac{x^3}{3}\right]_0^1 = \frac{1}{3}$ 2.-

Applying Stolz-Cesaro theorem $\displaystyle \lim_{n \to \infty} \frac{1}{n} \cdot \sum_{m = 1}^n \frac{m^2}{n^2} = \lim_{n \to \infty} \frac{(n + 1)^2}{(n + 1)^3 - n^3} = \frac{1}{3}$