Answer
This geometric series diverges.
Work Step by Step
Since the given series is in a similar form as the definition of the geometric series, we can make it very close by taking one 3 outside, i.e. $3^{n+1}$ becomes $3^{n}$.
The summation of an infinite geometric series is $S= \frac{a}{1-r}$ if |$r$| $\lt 1 $ (here $r$ is the common ratio of the geometric series).
Now, If we choose n = 0 we can see that the initial term is $a$ = $3$ and the common ratio is $r$ = $\frac{-3}{2}$ (just by looking at the form in the definition and the one in the question after factoring out a $3$).
$r$ = |$\frac{-3}{2}$| $\gt 1 $ , which means the series is divergent.
