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$$\require{cancel} \sum_\limits{n=1}^\infty \dfrac{(-2)^n}{n^2}=\sum_\limits{n=1}^{\infty} \dfrac{(-1)^n \ (2)^n}{n^2}$$

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?

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    $\begingroup$ This series is definitely divergent as you correctly concluded from other tests. So your question at the end is a bit unclear: what do you mean by “this is the alternating series test”? That test is inconclusive here because its conditions are not satisfied (the terms are alternating but they do not go to zero). $\endgroup$
    – Math101
    Commented Dec 30, 2019 at 22:07

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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.

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