Answer
Converges absolutely.
Work Step by Step
Consider $a_n=(-1)^n \dfrac{\sin n}{n^2}$
Here, $a_n$ is positive for all the values of $n$.
and $|a_n|=\dfrac{\sin n}{n^2}$
we see that $ \Sigma_{n=1}^\infty x\dfrac{\sin n}{n^2}\leq \Sigma_{n=1}^\infty\dfrac{1}{n^2}$ (a convergent p-series)
Thus, the series Converges absolutely.