Step 1: Write the general term of the series of absolute values as $b_n = \dfrac{1}{n^2(n+1)}$ and decompose it using partial fractions: $b_n = \dfrac{1}{n^2} + \left(\dfrac{1}{n+1} - \dfrac{1}{n}\right)$.
Step 2: Sum over $n=1$ to $N$: $\sum_{n=1}^N b_n = \sum_{n=1}^N \dfrac{1}{n^2} + \sum_{n=1}^N \left(\dfrac{1}{n+1}-\dfrac{1}{n}\right)$. The second sum telescopes to $\dfrac{1}{N+1} - 1$.
Step 3: Taking $N \to \infty$, $\sum_{n=1}^\infty \dfrac{1}{n^2} = \dfrac{\pi^2}{6}$ (a known convergent value) and the telescoped part tends to $-1$. So $\sum_{n=1}^\infty b_n = \dfrac{\pi^2}{6} - 1$, a finite number.
Step 4: Since the sum of absolute values is finite, the original alternating series converges absolutely, not just conditionally.
\[\boxed{\text{The series is absolutely convergent}}\]