Answer
Diverges
Work Step by Step
We have: $a_k=\dfrac{k-1}{k^2}$ and $b_k=\dfrac{1}{k}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{k-1}{k^2}}{1/k}\\=\lim\limits_{k \to \infty}\dfrac{k(k-1)}{k^2}=1$
Because $0 \lt \lim\limits_{k \to \infty}\dfrac{a_k}{b_k} \lt \infty $ and $\Sigma b_k$ is a harmonic divergent series.
Therefore, the series diverges by the limit comparison test
