<s>[INST] Is this a Riemann sum (if so, I can't figure out which one)? 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? 
 [/INST] $$\sum_{k=3}^n \frac{3}{k^2-k-2} = \sum_{k=3}^n \frac{1}{k-2}- \sum_{k=3}^n \frac{1}{k+1}$$
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 $n\geq 3$.
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