Answer
converges
Work Step by Step
We have: $a_k=\dfrac{1}{k^2-1}$ and $b_k=\dfrac{1}{k^2}$
Now, we need to apply the limit comparison test.
$L=\lim\limits_{k \to \infty}\dfrac{a_k}{b_k}\\=\lim\limits_{k \to \infty}\dfrac{\dfrac{1}{k^2-2}}{1/k^2}\\=\lim\limits_{k \to \infty} \dfrac{1}{1-\dfrac{1}{k^2}}\\=\lim\limits_{k \to \infty} \dfrac{1}{1-0} \\=1$
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 p-series with $p=2 \gt 1$.Therefore, the given series absolutely converges by the limit comparison test.
