<s>[INST] 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 $\frac{1}{2}$.
 [/INST] Let $\dfrac{2-r}{r(r+1)(r+2)}=f(r)-f(r+1),$ where$ f(m)=\dfrac{am+b}{m(m+1)}$
$\implies2-r=(r+2)(ar+b)-r(ar+a+b)=ar+2b$
Set $2b=2\iff b=1,ar=-r\iff a=-1$
$$\implies\sum_{r=1}^n\dfrac{2-r}{r(r+1)(r+2)}=\sum_{r=1}^n\left(f(r)-f(r+1)\right)=f(1)-f(n+1)$$
Set $n\to\infty$ and show that $n\to\infty f(n)=0$
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