Answer
Converges
Work Step by Step
Given $$\sum_{n=2}^{\infty} \frac{4 n^{2}+15 n}{3 n^{4}-5 n^{2}-17}$$
Compare with $\sum \frac{4}{3n^2}$, which is a convergent p-series with $p>1$. Using the limit comparison test gives:
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{4 n^{2}+15 n}{3 n^{4}-5 n^{2}-17} \cdot \frac{3 n^{2}}{4}\\
&=\lim _{n \rightarrow \infty} \frac{12 n^{4}+45 n^{3}}{12 n^{4}-20 n^{2}-68}\\
&=\lim _{n \rightarrow \infty} \frac{12+45 / n}{12-20 / n^{2}-68 / n^{4}}\\
&=1
\end{align*}
Thus $\sum_{n=2}^{\infty} \frac{4 n^{2}+15 n}{3 n^{4}-5 n^{2}-17} $ also converges.
