Looks like the series is
[tex]\displaystyle\sum_{n=1}^\infty \frac{12}{n(n+2)}[/tex]
Split up the summand into partial fractions:
[tex]\dfrac{12}{n(n+2)}=\dfrac6n-\dfrac6{n+2}[/tex]
The series has kth partial sum
[tex]S_k=\displaystyle\sum_{n=1}^k\frac{12}{n(n+2)}[/tex]
[tex]S_k=\left(6-2\right)+\left(3-\dfrac32\right)+\left(2-\dfrac65\right)+\left(\dfrac32-1\right)+\cdots+\left(\dfrac6{k-2}-\dfrac6k\right)+\left(\left(\dfrac6{k-1}-\dfrac6{k+1}\right)+\left(\dfrac6k-\dfrac6{k+2}\right)[/tex]
Several intermediate terms cancel to leave us with
[tex]S_k=6+3-\dfrac6{k+1}-\dfrac6{k+2}[/tex]
As [tex]k\to\infty[/tex], the last two terms converge to 0, so that
[tex]\displaystyle\lim_{k\to\infty}S_k=\sum_{n=1}^\infty\frac{12}{n(n+2)}=\boxed{9}[/tex]
