Answer
converges
Work Step by Step
Given $$\sum_{n=1}^{\infty} \frac{2^{n}+n}{3^{n}-2},\ \ \ \ b_{n}=\left(\frac{2}{3}\right)^{n}$$
We use the limit comparison test with $b_n$ given above:
\begin{align*}
\lim _{n \rightarrow \infty} \frac{a_{n}}{b_{n}}&=\lim _{n \rightarrow \infty} \frac{2^{n}+n}{3^{n}-2} \cdot \frac{3^{n}}{2^{n}}\\
&=\lim _{n \rightarrow \infty} \frac{6^{n}+n 3^{n}}{6^{n}-2^{n+1}}\\
&=\lim _{n \rightarrow \infty} \frac{1+n\left(\frac{1}{2}\right)^{n}}{1-2\left(\frac{1}{3}\right)^{n}}\\
&=1
\end{align*}
Since $\sum \left(\frac{2}{3}\right)^{n}$ is a geometric convergent series, then $\sum_{n=1}^{\infty} \frac{2^{n}+n}{3^{n}-2}$ also converges.
