limit of limits

We can factor out the $\frac{1}{n^{3}}$ in the sum and be left with $\frac{1}{n^{3}}(1^{2}+2^{2}+3^{2}+\ldots+n^{2})$ . The formula for the sum of consecutive squares up to $n$ is $\frac{n(n+1)(2n+1)}{6}$ . Thus, we can rewrite the limit:

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

Hence, $\frac{m}{n} = \frac{1}{3}$ and thus $m-n = 1-3 = \boxed{-2}$ .