How to find $\lim_{n\to\infty}\frac{1^2+2^2+...+n^2}{n^3-n+13}$ I need to find a limit of this sequence
 $$\lim_{n\to\infty}\frac{1^2+2^2+...+n^2}{n^3-n+13}$$
I've tried to use the squeeze theorem, but I could not find two sequences with the same limit. From Wolfram Alpha I learned that the limit is $\frac{1}{3}$, but I just can't find a way to get a sequence greater or less than this one, that would have this limit.
 A: HINT:
$$\sum_{k=1}^nk^2=\frac{n(n+1)(2n+1)}6\;.$$
A: L'Hopital it... For sequences the equivalent of L'Hopital rule is the Stolz Cezaro Theorem, which is this case says:
$$\lim_{n\to\infty}\frac{1^2+2^2+...+n^2}{n^3-n+13}=\lim_{n\to\infty}\frac{\left(1^2+2^2+...+n^2+(n+1)^2\right)-\left(1^2+2^2+...+n^2\right)}{\left((n+1)^3-(n+1)+13\right)-\left(n^3-n+13\right)}\\=\lim_{n\to\infty}\frac{(n+1)^2}{(n+1)^3-n^3+1}\\$$
Now both the denominator and numerator are quadratic polynomials.
A: Following Brian's solution, you expand for a cubic. Now think about this: As x reaches infinity, the terms don't really matter except for the x^3 terms. So the answer is...
A: An alternative - more general - method uses the fact that 
$\sum_{k=1}^{n}k^p=\frac{1}{p+1}n^{p+1}+P(n),$ in which $P(n)$ is a polynomial of degree $p$. This can be shown by induction.
For this question $p=2$, hence
$\frac{\sum_{k=1}^{n}k^2}{n^3-n+13}=\frac{\frac{1}{3}n^{3}+P(n)}{n^3-n+13}=\frac{\frac{1}{3}+P(n)/n^3}{1-n/n^3+13/n^3}.$
As $P(n)$ is a polynomial of degree 2, the limit $P(n)/n^3 \to 0$ for $n\to \infty$, hence the limit is $1/3$.
