If this is a telescoping series then how does it collapse? $\frac{3r+1}{r(r-1)(r+1)}$ Express $$\frac{3r+1}{r(r-1)(r+1)}$$ in partial fractions. Hence, or otherwise, show 
$$\sum_{r=2}^n\frac{3r+1}{r(r-1)(r+1)}=\frac52-\frac2n-\frac{1}{n+1}$$
So, I have obtained the partial fractions $$\frac{3r+1}{r(r-1)(r+1)}=\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}$$
then for $r=(2,3,4,5...(n-1),n)$
$$
\begin{align}
& {}+2-\frac12-\frac13 \\[8pt]
& {}+1-\frac13-\frac14 \\[8pt]
& {}+\frac23-\frac14-\frac15 \\[8pt]
& {}+\frac12-\frac15-\frac16 \\[8pt]
& {}+\cdots \\[8pt]
& {}+\frac{2}{n-2}-\frac{1}{n-1}-\frac{1}{n} \\[8pt]
& {}+\frac{2}{n-1}-\frac{1}{n}-\frac{1}{n+1}
\end{align}
$$
If this is a telescopic series how does it collapse?

HINT:
There are at least two methods for  $$\frac{3r+1}{r(r-1)(r+1)}=\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}$$
Method $1:$
$$\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}=-\left(\frac1{r+1}-\frac1r\right)-2\left(\frac1r-\frac1{r-1}\right)$$
The survivor of the summation $2\le r\le n$  after cancellation,
for the first part will be $\displaystyle-\left(\frac1{n+1}-\frac12\right)=\frac12-\frac1{n+1}$
for the second part will be $\displaystyle-2\left(\frac1n-\frac11\right)=2-\frac2n$
Method $2:$
$$\frac{2}{r-1}-\frac{1}{r}-\frac{1}{r+1}=\left(\frac1{r-1}-\frac1r\right)-\left(\frac1{r+1}-\frac1{r-1}\right)$$
Put a few values of $r$ like $2,3,4,5$ 
and $n-3,n-2,n-1,n$ to recognize the Telescopic nature & the surviving terms after cancellation 
