Answer
Converges
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 $a_k=\dfrac{\ln k}{k^2}$
1) In the given sequence, $a_k=\dfrac{\ln k}{k^2}$ , and $a_{k+1}=\dfrac{\ln (k+1)}{(k+1)^2} $ and $\dfrac{\ln (k+1)}{(k+1)^2} \leq \dfrac{\ln k}{k^2}$
This implies that $a_{k+1} \lt a_k$ and $a_{k+1} \gt 0$ , so the terms are a non-increasing sequence.
2) $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} \dfrac{\ln k}{k^2}= \lim\limits_{k \to \infty} \dfrac{1}{2k^2}=0$
This means that the given sequence converges.
