Evaluate $\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$ 
Evaluate
  $$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)}$$

So in a previous part of the question I calculated that 
$$\sum_{r=1}^{n} \frac{2-r}{r(r+1)(r+2)} = \sum_{r=1}^{n}\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\frac{n}{(n+1)(n+2)}$$
So my question is then why does $$\sum_{r=2}^{\infty} \frac{2-r}{r(r+1)(r+2)} = -\frac{1}{6}$$
As I though that it would be $\frac{1}{2}$.

Note that\begin{align}\frac{2-r}{r(r+1)(r+2)}&=\frac1r-\frac3{r+1}+\frac2{r+2}\\&=\frac1r-\frac1{r+1}-\frac2{r+1}+\frac2{r+2}.\end{align}Therefore, your series is naturally the sum of two telescoping series and its sum is$$\frac12-\frac23=-\frac16.$$
