This was supposedly an easy limit, and it is suspiciously similar to a Riemann sum, but I can't quite figure out for what function.
$$\lim_{n\to\infty}{\frac{1}{n} {\sum_{k=3}^{n}{\frac{3}{k^2-k-2}}}}$$
Well, even the fact that $\frac{3}{k^2-k-2} = \frac{1}{k-1}-\frac{1}{k+2}$ doesn't seem to simplify the problem. I thought this would be a telescoping sum, but it's clearly not.
Is that a Riemann sum at all?
On
$$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=3}^n \frac{1}{k-2}- \sum_{k=3}^n \frac{1}{k+1}$$
You (and I) were mistaken before, see @Romeo 's answer.
Notice that $$\sum_{k=3}^n \frac{1}{k-2}=\sum_{k=0}^{n-3} \frac{1}{k+1}$$
Insert above you get $$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=0}^{n-3} \frac{1}{k+1} - \sum_{k=3}^n \frac{1}{k+1} = \sum_{k=0}^2 \frac{1}{k+1} - \sum_{k=n-2}^{n} \frac{1}{k+1}= $$$$=1+\frac{1}{2}+\frac{1}{3} - \frac{1}{n-1}-\frac{1}{n}-\frac{1}{n+1}$$
Of course this argument requires $n\geq 3$.
Isn't it $k^2 -k -2 = (k+1)(k-2)$? In that way, $$ \frac{3}{k^2-k-2} = \frac{1}{k-2} - \frac{1}{k+1} $$ and this is likely to be telescopic.