Proof of the limit of a sequence The sequence is: 

$a_n$ = $\frac 1n$ [$(\frac 1n)^2 + (\frac 2n)^2 + (\frac 3n)^2...(\frac nn)^2$]

The objective is a proof of the limit from 1 to infinity. Just from toying around with a few examples, I speculate the limit is (1/3), but I'm not sure. This is how far I am into the proof (basically just the skeleton).

Let $\epsilon$ > 0. Choose N> _. Let n$\in \mathbb{N}$, n>N. We need |$\frac 1n$ [$(\frac 1n)^2 + (\frac 2n)^2 + (\frac 3n)^2...(\frac nn)^2$] - $\frac 13$| &lt; $\epsilon$. 

I don't really know how to manipulate this to get something useful. Any advice?
Hint: $$1^2+2^2+\ldots+n^2 = \dfrac{n(n+1)(2n+1)}{6}.$$
$$1+4+9+...+n^2=n(n+1)(2n+1)/6$$

so 

$$|1/n [(1/n)^2+(2/n)^2+(3/n)^2...(n/n)^2] - 1/3|=
|1/n^3 [(1)^2+(2)^2+(3)^2...(n)^2] - 1/3|$$

$$|1/n^3\cdot(n(n+1)(2n+1)/6) -1/3|=|(n+1)(2n+1)/6n^2 -1/3|=|(2n^2+3n+1-2n^2)/6n^2|$$

Can you go on?