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?
 A: 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 
A: Note that $\cfrac 2{r-1}=\cfrac 1{r-1}+\cfrac 1{r-1}$. Except at the beginning and the end one of these fractions will cancel with an element of the previous term, and one will cancel with an element of the term before that.
In your calculation you have $+1$ in the third row, which cancels with $-\frac 12$ in each of the first and second rows. And in the fourth row $\frac 23$ cancels with $-\frac13$ in the second and third rows.
A: HINT: $$\begin{align*}
+&2\color{black}{-\frac12}\color{red}{-\frac13}\\
+&\color{black}{1}\color{red}{-\frac13}\color{green}{-\frac14}\\
+&\color{red}{\frac23}\color{green}{-\frac14}\color{magenta}{-\frac15}\\
+&\color{green}{\frac12}\color{magenta}{-\frac15}-\frac16\\\\
+&\cdots
\end{align*}$$
A: Here is a breakdown into a single telescoping series.
$$
\begin{align}
\frac{3r+1}{r(r-1)(r+1)}
&=\frac2{r(r-1)}+\frac1{(r+1)r}\\
&=\left(\frac2{r-1}-\frac2r\right)+\left(\frac1r-\frac1{r+1}\right)\\
&=\underbrace{\left(\frac2{r-1}+\frac1r\right)}_{\large a_r}-\underbrace{\left(\frac2r+\frac1{r+1}\right)}_{\large a_{r+1}}
\end{align}
$$
Thus,
$$
\begin{align}
\sum_{r=2}^n\frac{3r+1}{r(r-1)(r+1)}
&=\sum_{r=2}^n(a_r-a_{r+1})\\
&=a_2-a_{n+1}\\[3pt]
&=\frac52-\left(\frac2n+\frac1{n+1}\right)
\end{align}
$$
