Answer
Divergent
Work Step by Step
Let $a_n=\frac{(-2)^n}{n^2}$. Then $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|= \lim\limits_{n \to \infty}\left|\frac{\frac{(-2)^{n+1}}{(n+1)^2}}{\frac{(-2)^n}{n^2}}\right|=\lim\limits_{n \to \infty}\left|\frac{(-2)^{n+1}}{(n+1)^2}\cdot\frac{n^2}{(-2)^n}\right|\lim\limits_{n \to \infty}\left|\frac{(-2)^{n+1}}{(-2)^n}\cdot\frac{n^2}{(n+1)^2}\right|=\lim\limits_{n \to \infty}\left|-2\cdot\frac{n^2}{(n+1)^2}\right| =$
$$2\cdot\lim\limits_{n \to \infty}\frac{n^2}{(n+1)^2} =2\cdot 1 =2$$ since the numerator and denominator in the last limit have the same degree.
Since $\lim\limits_{n \to \infty}\left|\frac{a_{n+1}}{a_n}\right|=2 >1$, the series $\displaystyle{\sum_{n=1}^{\infty}\frac{(-2)^n}{n^2}}$ is divergent by the Ratio Test.