Answer
Diverges.
Work Step by Step
For alternating series $\Sigma_{k=1}^\infty(-1)^k a_k$ (let us consider), to be convergent , we must follow the two conditions such as: a) The magnitude of terms must form a non-increasing sequence.
b) $\lim\limits_{k \to \infty} a_k=0$
We are given that $a_k=\dfrac{k^2-1}{k^2+3}$
In the given sequence, $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \dfrac{k^2-1}{k^2+3}\\ =\lim\limits_{k \to \infty} \dfrac{1-1/k^2}{1+3/k^2}\\=\dfrac{1-0}{1+0}\\= 1 \ne 0$
This implies that the second condition for a alternating series does not satisty.
Thus, the given sequence diverges.
