$$\require{cancel}
\sum_\limits{n=1}^\infty \dfrac{(-2)^n}{n^2}=\sum_\limits{n=1}^{\infty} \dfrac{(-1)^n \ (2)^n}{n^2}$$
According to the alternating series test which states that the series terms are eventually decreasing.
Alternating Series Test
$$\sum_\limits{n=1}^\infty \dfrac{(-2)^n}{n^2}=-2+1-\frac{8}9+1\ \ …$$
I noticed that for the first condition of the alternating series test was that the series kept on increasing for all $n$.Then I thought about the doing the following:
$$\lim_\limits{n \to \infty} \dfrac{2^n}{n^2}=\lim_\limits{n \to \infty} \dfrac{\ln(2) \ 2^n}{2n}= \lim_\limits{n \to \infty} \dfrac{\ln^2(2) \ 2^n}{2}=\infty \ \mathbf{or \ divergent}$$
I am wondering the following since this is the Alternating series test I can not assert that the series is divergent right? Because under this assumption which I made which was that the test was inconclusive I did the Ratio Test.
Ratio Test
$$\sum_\limits{n=1}^\infty \vert \dfrac{(-2)^n}{n^2} \vert = \sum_\limits{n=1}^\infty \frac{2^n}{n^2} $$
$$\lim_\limits{n \to \infty} \frac{\frac{2^n \ \cdot \ 2}{(n+1)^2}}{\frac{2^n}{n^2}}=\lim_\limits{n \to \infty} \frac{\frac{\cancel{2^n} \ \cdot \ 2}{(n+1)^2}}{\frac{\cancel{2^n}}{n^2}}=\lim_\limits{n \to \infty} \dfrac{2n^2}{n^2+2n+1} = 2$$
Using the Ratio Test I reached the conclusion it was divergent.
Reminder the Question:
I am wondering the following since this is the Alternating series test I can not assert that the series is divergent right?
Best Answer
Yes, the alternate series test is inconclusive here. But there is no need to apply the ratio test. Since you don't have $\lim_{n\to\infty}\frac{(-2)^n}{n^2}=0$, the series diverges.