Answer
Conditionally convergent.
Work Step by Step
We have $\mathop \sum \limits_{k = 1}^\infty \dfrac{{k\cos k\pi }}{{{k^2} + 1}}$.
For $k \ge 1$, we have $\cos k\pi = {\left( { - 1} \right)^k}$. Thus, the series can be rewritten as $\mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^k}k}}{{{k^2} + 1}}$, where ${a_k} = \dfrac{k}{{{k^2} + 1}}$.
1. The series is decreasing since ${a_k} \gt {a_{k + 1}}$ for $k \ge 1$.
Since
$\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{k}{{{k^2} + 1}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{k + \dfrac{1}{k}}} = 0$,
by the alternating series test, the series converges.
2. Now, we test the series for absolute convergence by examining the series of absolute values:
$\mathop \sum \limits_{k = 1}^\infty \left| {\dfrac{{k\cos k\pi }}{{{k^2} + 1}}} \right| = \mathop \sum \limits_{k = 1}^\infty \dfrac{k}{{{k^2} + 1}}$
The leading term is $\dfrac{k}{{{k^2}}} = \dfrac{1}{k}$. Thus, it suggests that we examine the convergence by using the Limit Comparison Test with the series of $\dfrac{1}{k}$.
Let ${b_k} = \dfrac{1}{k}$. Evaluate:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{a_k}}}{{{b_k}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{k^2}}}{{{k^2} + 1}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{1 + \dfrac{1}{{{k^2}}}}} = 1$
We obtain $\rho = 1 \gt 0$. Since the harmonic series $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{k}$ diverges, by the Limit Comparison Test, the series $\mathop \sum \limits_{k = 1}^\infty \dfrac{k}{{{k^2} + 1}}$ diverges. Therefore, the original series diverges absolutely.
Since the original series converges and also diverges absolutely, it converges conditionally.
