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 $\frac{1}{2}$.

If $H_{n}$ is the $n$-th harmonic number, we have $$\sum_{r=2}^{n}\left(\frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)=\left(H_{n}-1\right)-3\left(H_{n}-1-\frac{1}{2}\right)+2\left(H_{n}-1-\frac{1}{2}-\frac{1}{3}\right)$$ $$=-1+\frac{3}{2}-\frac{2}{3}=-\frac{1}{6}.$$ Now take the limit as $n\rightarrow+\infty$.
