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$$\frac12-\frac23=-\frac16.$$
Note that \begin{align}\sum_{r=2}^\infty\left( \frac{1}{r}-\frac{3}{r+1}+\frac{2}{r+2}\right)&=\sum_{r=2}^\infty\left(\frac1r-\frac1{r+1}\right)-2\sum_{r=2}^\infty\left(\frac1{r+1}-\frac1{r+2}\right)\\&=\left(\frac12-\frac13+\frac13-\frac14+\cdots\right)-2\left(\frac13+\frac14-\frac14+\frac15-\cdots\right)\\&=\frac12-\frac23=-\frac16\end{align}
Actually your finite sum is equal to $$\sum_{r=2}^n\frac{2-r}{r(r+1)(r+2)}=\frac{2}{n+2}-\frac{1}{n+1}-\frac{1}{6}$$
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 $n\to\infty f(n)=0$
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 $n\rightarrow+\infty$.
By definition,
$$ \sum_{r=1}^\infty a_r = \lim_{n\to\infty}\sum_{r=1}^n a_r .$$
In your case the above limit is zero. Then:
$$ \sum_{r=2}^\infty a_r = \sum_{r=1}^\infty a_r - a_1 = -a_1 . $$
This reasoning is what @HenriLee was referring to in his answer.
