Answer
Converges
Work Step by Step
We can see that $\Sigma_{k=1}^{\infty} \dfrac{4}{2+3^k k} \lt \Sigma_{k=1}^{\infty} \dfrac{4}{3^k k} \lt \Sigma_{k=1}^{\infty} \dfrac{4}{3^k }$
But the series $\Sigma_{k=1}^{\infty} \dfrac{4}{3^k }$ is a geometric series with a common ratio of $\dfrac{1}{3}$, so it converges by the direct limit. This implies that the given series is less than a converging series, so it converges.
