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