Answer
Converges
Work Step by Step
Let $u_n=(\dfrac{n}{3n+1})^n$
When the numerator increases, the value of fraction is always increasing and also, when the denominator decreases, the value of fraction is always increasing.
$\implies u_n \leq (\dfrac{n}{3n})^n=(\dfrac{1}{3})^n$
Here, the series $\Sigma_{n=1}^\infty (\dfrac{1}{3})^n$ shows a convergent series by the comparison test.