Answer
Converges
Work Step by Step
We have: $a_k=\dfrac{3}{e^k}$ and $b_k=\dfrac{3}{2+e^k}$
We see that $\dfrac{3}{2+e^k} \leq \dfrac{3}{e^k}$
But the series $\Sigma \dfrac{1}{e^k}$ converges because it shows a geometric series with $r=\dfrac{1}{e} \lt 1$
Therefore, the given series converges by the comparison test.
