Answer
Converges absolutely
Work Step by Step
Let us consider $a_n=(-1)^{n+1}\dfrac{n}{n^3+1}$
and $|a_n|=\dfrac{n}{n^3+1}$
Apply the limit comparison test.
This implies that $\Sigma_{n=1}^\infty\dfrac{n}{n^3+1} \leq \Sigma_{n=1}^\infty\dfrac{n}{n^3}=\Sigma_{n=1}^\infty\dfrac{1}{n^2}$
We can see that $\Sigma_{n=1}^\infty\dfrac{1}{n^2}$ shows a p-series with $p=2$ .
Hence, the series converges absolutely.
