Answer
The series diverges.
Work Step by Step
Apply the ratio test.
$$\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_{n}}$$
If $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_{n}} = L \lt 1$, then the series is absolutely convergent.
If $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_{n}} = L \gt 1$ or $L=\infty$, then the series is divergent.
If $\lim\limits_{n \to \infty} \frac{a_{n+1}}{a_{n}} = L = 1$, then the Ratio Test is inconclusive, so no conclusion can be drawn about the series' convergence.
Plugging in values, we get:
$\lim\limits_{n \to \infty} \frac{3^{n+1}}{2^{n+1}(n+1)^{3}} \times \frac{2^{n}n^{3}}{3^{n}}$
After simplifying exponents, we get:
$\lim\limits_{n \to \infty} \frac{3n^{3}}{2^{n}(n+1)^{3}}$
Expanding the denominator and evaluating, it results in:
$\lim\limits_{n \to \infty} \frac{3n^{3}}{2(n^{3}+3n^{2}+3n+1)}$ = $\frac{3}{2} \gt 1$
Thus, by the Ratio Test, the series diverges.
