Answer
Diverges by Th. 9.12.
Work Step by Step
Th. 9.12 Direct Comparison Test:
Let $0 < a_{n} \leq b_{n}$ for all $n$.
1. If $\displaystyle \sum_{n=1}^{\infty}b_{n}$ converges, then $\displaystyle \sum_{n=1}^{\infty}a_{n}$ converges.
2. If $\displaystyle \sum_{n=1}^{\infty}a_{n}$ diverges, then $\displaystyle \sum_{n=1}^{\infty}b_{n}$ diverges.
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$\displaystyle \sum_{n=1}^{\infty}(\frac{3}{2})^{n}$ is a divergent
geometric series $(p=\displaystyle \frac{3}{2} > 1)$, we aim for case (2).
Letting $a_{n}=(\displaystyle \frac{3}{2})^{n}$, we have
$b_{n}= \displaystyle \frac{3^{n}}{2^{n}-1} > (\frac{3}{2})^{n} > 0,$
$0 < a_{n} \leq b_{n}$, and
$\displaystyle \sum_{n=1}^{\infty}a_{n}=\sum_{n=1}^{\infty}(\frac{3}{2})^{n}$ diverges , so
$\displaystyle \sum_{n=1}^{\infty}b_{n}=\sum_{n=1}^{\infty}\frac{3^{n}}{2^{n}-1}$ also diverges.