$\sum_{k=1}^{n} (k^2-k)(n-k) - 1$
$= \sum (k^2-k)n - \sum (k^2-k)k -1$
$= n\sum (k^2-k) - \sum (k^3-k^2) -1$
$= n\left(\sum k^2 - \sum k\right) - \left(\sum k^3 - \sum k^2\right) -1$

Denominator: $\sum k^3 - \sum k^2$

So expression becomes: $\frac{n(\sum k^2 - \sum k) - (\sum k^3 - \sum k^2)}{\sum k^3 - \sum k^2}$

For large $n$, use leading terms: $\sum k \sim \frac{n^2}{2}, \quad \sum k^2 \sim \frac{n^3}{3}, \quad \sum k^3 \sim \frac{n^4}{4}$

Now: Numerator leading term: $= n\left(\frac{n^3}{3} - \frac{n^2}{2}\right) - \left(\frac{n^4}{4} - \frac{n^3}{3}\right)$
$= \frac{n^4}{3} - \frac{n^3}{2} - \frac{n^4}{4} + \frac{n^3}{3}$
$= \left(\frac{1}{3}-\frac{1}{4}\right)n^4 + \left(-\frac{1}{2}+\frac{1}{3}\right)n^3$
$= \frac{1}{12}n^4 - \frac{1}{6}n^3$

Denominator: $= \frac{n^4}{4} - \frac{n^3}{3}$

Taking ratio (dominant terms): $= \frac{\frac{1}{12}n^4}{\frac{1}{4}n^4} = \frac{1/12}{1/4} = \frac{1}{3}$

Final Answer: $ \boxed{\frac{1}{3}} $