I assume the series is supposed to be
[tex]\displaystyle\sum_{n=1}^\infty \frac6{n(n+3)}[/tex]
The summand can be expanded into partial fractions:
[tex]\dfrac6{n(n+3)}=\dfrac an+\dfrac b{n+3}[/tex]
[tex]\implies 6=a(n+3)+bn=(a+b)n+3a[/tex]
[tex]\implies\begin{cases}a+b=0\\3a=6\end{cases}\implies a=2,b=-2[/tex]
Then the sum is equivalent to
[tex]\displaystyle2\sum_{n=1}^\infty\left(\frac1n-\frac1{n+3}\right)[/tex]
Consider the N-th partial sum of the series:
2[(1 - 1/4) + (1/2 - 1/5) + (1/3 - 1/6) + (1/4 - 1/7) + (1/5 - 1/8) + …
+ (1/(N - 3) - 1/N) + (1/(N - 2) - 1/(N + 1)) + (1/(N - 1) - 1/(N + 2)) + (1/N - 1/(N + 3))]
Each of the negative terms (in bold) will have a positive counterpart (underlined) that cancel, so the N-th partial sum would be
[tex]\displaystyle\sum_{n=1}^N\frac6{n(n+3)} = 2\left(1+\frac12+\frac13-\frac1{N+1}-\frac1{N+2}-\frac1{N+3}\right)[/tex]
As N approaches infinity, the last three terms in the sum converge to 0, so the original sum converges to
[tex]\displaystyle\sum_{n=1}^\infty \frac6{n(n+3)} = 2\left(1+\frac12+\frac13\right) = \boxed{\frac{11}3}[/tex]
