Answer
convergent
using Limit Comparison Test
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{3 }{n(n+3)}$$
Use the Limit Comparison Test with $a_n =\dfrac{3 }{n(n+3)}$ and $b_n=\dfrac{ 1}{ n^{2}}$
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{3n^2 }{n(n+3)}\\
&=\lim _{n \rightarrow \infty} \frac{3 }{ (1+3/n)}\\
&=3
\end{align*}
Since $\displaystyle \sum_{n=1}^{\infty} \frac{1}{ n^2}$ is convergent ($p-$series series $p>1$) , then $\displaystyle\sum_{n=1}^{\infty} \frac{3 }{n(n+3)}$ is also convergent