<s>[INST] 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?
 [/INST] 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,
$$
\begin{align}
\sum_{r=2}^n\frac{3r+1}{r(r-1)(r+1)}
&=\sum_{r=2}^n(a_r-a_{r+1})\\
&=a_2-a_{n+1}\\[3pt]
&=\frac52-\left(\frac2n+\frac1{n+1}\right)
\end{align}
$$
</s>