Answer: \(\displaystyle \frac{1}{r(r+1)(r-1)}=\frac{1}{2(r-1)}-\frac{1}{r}+\frac{1}{2(r+1)}\).
\(\displaystyle \sum_{r=2}^{n}\frac{1}{r(r+1)(r-1)}=\frac14-\frac{1}{2n}+\frac{1}{2(n+1)}\).
\(\displaystyle \sum_{r=2}^{\infty}\frac{1}{r(r+1)(r-1)}=\frac14\).
Write the partial fraction decomposition as
\(\frac{1}{r(r-1)(r+1)}=\frac{A}{r-1}+\frac{B}{r}+\frac{C}{r+1}.\)
Multiplying by \(r(r-1)(r+1)\) gives
\(1=Ar(r+1)+B(r-1)(r+1)+Cr(r-1).\)
Put \(r=1\): \(1=2A\), so \(A=\frac12\). Put \(r=0\): \(1=-B\), so \(B=-1\). Put \(r=-1\): \(1=2C\), so \(C=\frac12\). Hence
\(\frac{1}{r(r-1)(r+1)}=\frac{1}{2(r-1)}-\frac{1}{r}+\frac{1}{2(r+1)}.\)
Therefore
\(\sum_{r=2}^{n}\frac{1}{r(r-1)(r+1)}=\sum_{r=2}^{n}\left\{\frac{1}{2(r-1)}-\frac1r+\frac{1}{2(r+1)}\right\}.\)
Writing out the first and last terms shows the cancellation:
\(\left(\frac12-\frac12+\frac16\right)+\left(\frac14-\frac13+\frac18\right)+\cdots+\left(\frac{1}{2(n-1)}-\frac1n+\frac{1}{2(n+1)}\right).\)
All middle terms cancel, leaving
\(\sum_{r=2}^{n}\frac{1}{r(r-1)(r+1)}=\frac14-\frac{1}{2n}+\frac{1}{2(n+1)}.\)
As \(n\to\infty\), the terms \(-\frac{1}{2n}\) and \(\frac{1}{2(n+1)}\) tend to \(0\). Thus
\(\sum_{r=2}^{\infty}\frac{1}{r(r-1)(r+1)}=\frac14.\)
