Answer
Converges
Work Step by Step
We have: $a_k=\dfrac{2^k}{e^k-1}$ and $b_k=\dfrac{2^k}{e^k}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{2^k}{e^k-1}}{2^k/e^{k}}\\=\lim\limits_{k \to \infty}\dfrac{2^k e^k}{2^k(e^k-1)} \\=1$
Therefore, the series converges by the limit comparison test.
