Answer
Converges
Work Step by Step
Consider $a_n=\dfrac{1}{n3^{n}}$
When we increase the numerator, the value of the fraction will always increase and when we decrease the denominator, the value of the fraction will always increase.
Thus, we have $a_n \leq \dfrac{1}{3^{n}}$
Here, $\Sigma_{n=1}^\infty \dfrac{1}{3^n}$ is a convergent geometric series with common ratio $r=\dfrac{1}{3} \gt 1$.
Thus the series converges by the comparison test.