Confusion in this limits problem 
Evaluate $$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3}$$

This is the original method to solve this is:
Taking summation of the square numbers $1^2 + 2^2 + 3^2 +...+n^2 = \frac{1}{6}n(n+1)(2n+1)$
$$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3} = \frac{\frac{1}{6}n(n+1)(2n+1)}{n^3}$$
$$=\lim\limits_{n \to \infty} \frac{(n+1)(2n+1)}{6n^2} = \lim\limits_{n\to \infty} \frac{1}{6}(1+\frac{1}{n})(2+\frac{1}{n})$$
$$=\frac{2}{6} = \frac{1}{3}$$
But when looking at the limit in a different angle I get a different answer,
$$\lim\limits_{n\to \infty} \frac{1^2 + 2^2 +...+n^2}{n^3} = \lim\limits_{n\to \infty} \frac{1^2}{n^3}+\frac{2^2}{n^3}+...+\frac{1}{n}$$
$$=0+0+...+0 = 0$$
Both the method seem right to me, but why I am getting different answers? What have I done wrong? Please Explain. Thank you!

Converting the limit into an integral is one right way to evaluate it. 
$$\lim_{n\to \infty}\dfrac{1^2+2^2+\cdots+n^2}{n^3}=\lim_{n\to \infty}\dfrac{1}{n}\sum_{k=1}^{n}\left(\dfrac{k}{n}\right)^2=\int_{0}^{1}x^2\mathrm dx$$
