Answer: The sum of the first \(n\) terms is \(\frac{1}{2}\left(\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)\), so the sum to infinity is \(\frac{3}{4}\).
Let
\(S_n=\displaystyle\sum_{r=1}^{n}\frac{1}{r(r+2)}\).
Use partial fractions:
\(\frac{1}{r(r+2)}=\frac{1}{2}\left(\frac{1}{r}-\frac{1}{r+2}\right)\).
So
\(S_n=\frac{1}{2}\sum_{r=1}^{n}\left(\frac{1}{r}-\frac{1}{r+2}\right)\).
Write out the terms:
\(S_n=\frac{1}{2}\left[\left(1-\frac{1}{3}\right)+\left(\frac{1}{2}-\frac{1}{4}\right)+\left(\frac{1}{3}-\frac{1}{5}\right)+\cdots+\left(\frac{1}{n}-\frac{1}{n+2}\right)\right]\).
Most terms cancel, leaving
\(S_n=\frac{1}{2}\left(1+\frac{1}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)\).
Hence
\(S_n=\frac{1}{2}\left(\frac{3}{2}-\frac{1}{n+1}-\frac{1}{n+2}\right)=\frac{3}{4}-\frac{1}{2(n+1)}-\frac{1}{2(n+2)}\).
As \(n\to\infty\), the last two terms tend to \(0\), so the sum to infinity is
\(\displaystyle \frac{3}{4}\).
