Answer
Converges
Work Step by Step
Let us consider an alternating series $\Sigma_{k=1}^\infty(-1)^k a_k$ that can be convergent , when it satisfy the following two conditions such as: 1) The magnitude of terms must form a non-increasing sequence.
2) $\lim\limits_{k \to \infty} a_k=0$
Here, in the problem we have $a_k=\dfrac{1}{k \ln ^2 k}$
a) In the given sequence, $a_k=\dfrac{1}{k \ln ^2 k}$ , and $a_{k+1}=\dfrac{1}{(k+1) \ln ^2 (k+1)}$ and $\dfrac{1}{(k+1) \ln ^2 (k+1)}\leq \dfrac{1}{k \ln ^2 k}$
This implies that $a_{k+1} \lt a_k$ and $a_{k+1} \gt 0$ , so the terms are a non-increasing sequence.
b) $\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} (\dfrac{1}{k \ln ^2 k})$
Apply L-Hospital's Rule to obtain:
$\lim\limits_{k \to \infty} a_k=\lim\limits_{k \to \infty} (\dfrac{-1/k^2}{2 \ln k/k})=\lim\limits_{k \to \infty} (\dfrac{1/k^2}{2/k})=\lim\limits_{k \to \infty} (\dfrac{1}{2k})=0$
This implies that the given series converges.
