I guess the series is
[tex]\displaystyle\sum_{n=3}^\infty\frac8{(n-1)(n+1)}[/tex]
Splitting into partial fractions gives
[tex]\dfrac8{(n-1)(n+1)}=\dfrac4{n-1}-\dfrac4{n+1}[/tex]
The series telescopes, which we can see by considering the [tex]k[/tex]-th partial sum:
[tex]\displaystyle\sum_{n=3}^k\frac8{(n-1)(n+1)}=(2-1)+\left(\frac43-\frac45\right)+\cdots+\left(\frac4{k-2}-\frac4k\right)+\left(\frac4{k-1}-\frac4{k+1}\right)[/tex]
(feel free to write out more terms to convince yourself that the following is true)
[tex]\displaystyle\sum_{n=3}^k\frac8{(n-1)(n+1)}=2+\frac43-\frac4k-\frac4{k+1}[/tex]
Then as [tex]k\to\infty[/tex], we end up with
[tex]\displaystyle\lim_{k\to\infty}\sum_{n=3}^k\frac8{(n-1)(n+1)}=\boxed{\frac{10}3}[/tex]
