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 $\frac{1}{2}$.

Actually your finite sum is equal to $$\sum_{r=2}^n\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}-\frac{1}{6}$$
