Answer
Converges
Work Step by Step
Given $$ \sum_{n=1}^{\infty} \frac{n^2+4n}{3n^4+9}$$
Compare with the convergent series $ \sum_{n=1}^{\infty} \frac{1}{n^2}$ ($p-$series, $p>1$); then by using the limit comparison test, we get:
\begin{align*}
\lim_{n\to \infty} \frac{a_n}{b_n}&=\lim_{n\to \infty}\frac{n^4+4n^3}{3n^4+9}\\
&=\lim_{n\to \infty}\frac{1+4/n}{3 +9/n^4}\\
&=\frac{1}{3}
\end{align*}
Thus the given series also converges.
