$\begin{align}  & \frac{1}{r(r+2)}=\frac{A}{r}+\frac{B}{r+2} \\ & 1=A(r+2)+Br \\ & Let\text{ }r=-2 \\ & 1=-2B \\ & B=-\frac{1}{2} \\ & when\text{ }r=0 \\ & 1=2A \\ & A=\frac{1}{2} \\ & \frac{1}{r(r+2)}=\frac{1}{2r}-\frac{1}{2(r+2)}=\frac{1}{2}\left( \frac{1}{r}-\frac{1}{r+2} \right) \\ & \therefore \sum\limits_{r=1}^{n}{\frac{1}{r\left( r+2 \right)}=\sum\limits_{r=1}^{n}{\frac{1}{2}\left[ \frac{1}{r}-\frac{1}{r+2} \right]}}=\frac{1}{2}\left[ \sum\limits_{r=1}^{n}{\frac{1}{r}-\sum\limits_{r=1}^{n}{\frac{1}{r+2}}} \right] \\ & \sum\limits_{r=1}^{n}{\frac{1}{r}}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+---+\text{ }\frac{1}{n} \\ & \sum\limits_{r=1}^{n}{\frac{1}{r+2}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+---+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}} \\ & \therefore \text{  }\sum\limits_{r=1}^{n}{\frac{1}{r\left( r+2 \right)}=\sum\limits_{r=1}^{n}{\frac{1}{2}\left[ \frac{1}{r}-\frac{1}{r+2} \right]}}=\text{ }\frac{1}{2}\left[ \sum\limits_{r=1}^{n}{\frac{1}{r}-\sum\limits_{r=1}^{n}{\frac{1}{r+2}}} \right] \\ & \sum\limits_{r=1}^{n}{\frac{1}{r}}=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+---+\frac{1}{n} \\ & \sum\limits_{r=1}^{n}{\frac{1}{r+2}=\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+---+\frac{1}{n}+\frac{1}{n+1}+\frac{1}{n+2}} \\ & Hence \sum\limits_{r=1}^{n}{\frac{1}{r\left( r+2 \right)}=\frac{1}{2}\left[ \frac{3}{2}-\frac{2n+3}{\left( n+1 \right)\left( n-2 \right)} \right]} \\ & \text{The sum to Infinity, we write} \\ & {{S}_{n}} = \frac{1}{2}\left[ \frac{3}{2}-\frac{2+3}{\left( 1+\frac{1}{n} \right)\left( 1-\frac{2}{n} \right)} \right]  \\ & \underset{n\to \infty }{\mathop{Lim{{S}_{n}}}}\,=\frac{1}{2}\left[ \frac{3}{2}-\frac{2}{1} \right] = \frac{1}{2}\left( \frac{-1}{2} \right)=\underline{\underline{\frac{-1}{4}}} \\\end{align}$