Answer
Conditionally convergent.
Work Step by Step
We have $\mathop \sum \limits_{k = 1}^\infty \dfrac{{\cos k\pi }}{k}$.
For $k \ge 1$, we have $\cos k\pi = {\left( { - 1} \right)^k}$. Thus, the series can be rewritten as $\mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{k}$, where ${a_k} = \dfrac{1}{k}$.
1. The series is decreasing since ${a_k} \gt {a_{k + 1}}$ for $k \ge 1$. Since $\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{k} = 0$, by the alternating series test, the series converges.
2. Now, we test the series for absolute convergence by examining the series of absolute values:
$\mathop \sum \limits_{k = 1}^\infty \left| {\dfrac{{{{\left( { - 1} \right)}^k}}}{k}} \right| = \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{k}$
The series on the right-hand side is a divergent harmonic series, hence the original series diverges absolutely.
Since the original series converges and also diverges absolutely, therefore it converges conditionally.
