Answer
Converges
Work Step by Step
We have: $a_k=\dfrac{2}{e^k-e^{-k}}$ and $b_k=\dfrac{1}{e^k}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{2}{e^k-e^{-k}}}{1/ e^k}\\=\lim\limits_{k \to \infty} \dfrac{2}{1-e^{-2k}} \\=\dfrac{2}{1-0}\\=2$
We see that $0 \lt \lim\limits_{k \to \infty}\dfrac{a_k}{b_k} \lt \infty$
and the series $\Sigma b_k$ is a convergent geometric series with $r=\dfrac{1}{e}$.Therefore, the given series converges by the limit comparison test.
