Answer
Converges absolutely
Work Step by Step
Consider $u_n=(-1)^{n+1}\dfrac{n}{n^3+1}$
Here, $u_n$ is positive for all the values of $n$.
and $|u_n|=\dfrac{n}{n^3+1}$
We see that
$\Sigma_{n=1}^\infty\dfrac{n}{n^3+1} \leq \Sigma_{n=1}^\infty\dfrac{n}{n^3}=\Sigma_{n=1}^\infty\dfrac{1}{n^2}$ (a convergent p-series)
Thus, the series converges absolutely.