## Calculus (3rd Edition)

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.