Answer
Converges Conditionally
Work Step by Step
Let us consider $a_n=(-1)^{n+1}\dfrac{1+n}{n^2}$
we can see that $\Sigma_{n=1}^\infty |a_n|=\Sigma_{n=1}^\infty \dfrac{1+n}{n^2}=\Sigma_{n=1}^\infty \dfrac{1}{n^2}+\Sigma_{n=1}^\infty \dfrac{1}{n}$
Here,we can see that the series $\Sigma_{n=1}^\infty \dfrac{1}{n^2}$ shows a convergent p-series and the series $\Sigma_{n=1}^\infty \dfrac{1}{n}$ shows a divergent p-series.
This implies that the sum of both the series defines divergent.
Need to consider $b_n=\dfrac{1}{n^2}+\dfrac{1}{n}$
we can see that $b_n$ is increasing and $\lim\limits_{n \to \infty} b_n=\lim\limits_{n \to \infty} (\dfrac{1}{n^2})+\lim\limits_{n \to \infty} (\dfrac{1}{n})=0$
Therefore, the series Converges Conditionally by the Alternating Series Test.
