Answer
Conditionally convergent.
Work Step by Step
We have $\mathop \sum \limits_{k = 1}^\infty \dfrac{{{{\left( { - 1} \right)}^k}}}{{\sqrt {k\left( {k + 1} \right)} }}$. Let ${a_k} = \dfrac{1}{{\sqrt {k\left( {k + 1} \right)} }}$.
1. The series is decreasing since ${a_k} \gt {a_{k + 1}}$ for $k \ge 1$. Since $\mathop {\lim }\limits_{k \to \infty } {a_k} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{\sqrt {k\left( {k + 1} \right)} }} = 0$, by the alternating series test, the series converges.
2. Now, we test the series for absolute convergence by examining the series of absolute values:
$\mathop \sum \limits_{k = 1}^\infty \left| {\dfrac{{{{\left( { - 1} \right)}^k}}}{{\sqrt {k\left( {k + 1} \right)} }}} \right| = \mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{\sqrt {k\left( {k + 1} \right)} }}$
The leading term is $\dfrac{1}{{\sqrt {{k^2}} }} = \dfrac{1}{k}$. 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:
$\rho = \mathop {\lim }\limits_{k \to \infty } \dfrac{{{a_k}}}{{{b_k}}} = \mathop {\lim }\limits_{k \to \infty } \dfrac{k}{{\sqrt {k\left( {k + 1} \right)} }} = \mathop {\lim }\limits_{k \to \infty } \dfrac{1}{{\sqrt {1 + \dfrac{1}{k}} }} = 1$
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 $\mathop \sum \limits_{k = 1}^\infty \dfrac{1}{{\sqrt {k\left( {k + 1} \right)} }}$ diverges. Hence, the original series diverges absolutely.
Since the original series converges and also diverges absolutely, it converges conditionally.
