Answer
Diverges
Work Step by Step
We have: $a_k=\dfrac{k^2-1}{k^3+4}$ and $b_k=\dfrac{1}{k}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{k^2-1}{k^3+4}}{1/k}\\=\lim\limits_{k \to \infty}\dfrac{k^3-k}{k^3+4}\\=1$
Therefore, the series diverges by the limit comparison test.
