Answer
Converges absolutely
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) $\lim\limits_{k \to \infty} a_k=0$
The series will converge conditionally when it is a convergent series , but the condition is when the series of its absolute value diverges and the series will converge absolutely but the condition is when the series of its absolute value converges.
The convergence of a p-series can be stated as when an infinite series let us say $\sum_{n=1}^{n=\infty} \dfrac{1}{n^p}$ converges when $p \gt 1$ and
and otherwise it will diverge.
Here, in the problem we have $\sum\dfrac{|\cos k|}{k^3} \leq \dfrac{1}{k^{3}}$
But the series $\sum \dfrac{1}{k^3}$ is a convergent p-series with $p=3$. This implies that the given series converges by p-test.
Thus, we can conclude that absolute value of series converges , so the given series converges absolutely.
