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# Use the Ratio Test to determine whether the series is convergent or divergent.$\displaystyle \sum_{n = 1}^{\infty} \frac {( - 2)^n}{n^2}$

## Diverges

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let's use the ratio test to determine whether this series converges or diverges. Now, the ratio test involves looking at this term here a n So in our case, this is negative two to the end over and squared. So we're interested in the answer to this limit. Here we look at absolute value a n plus one over a n and notice that the plus one, the one is being added to the n, not a n. So let's just go ahead and evaluate numerator and denominator here, so and plus one maybe do this in red, right? This is negative. Two and plus one right over n plus one square. Yeah, and then in blue and all right, there it is down there just using our formula up here for a M. Uh huh. And then now it's maybe clean this up a little bit by writing it as a product, and I could lose the absolute value. And when I do that, I'll get remember these negative signs here. So and I also I'll rewrite this as two times two to the end. Yeah, Now we see by doing this the reason for doing this trick up here is so that we could cancel the to to the ends. And now this is all equal to the limit and goes to infinity two and squared over and plus one square. This limit is of the form infinity over infinity. But you could do some algebra to simplify this. You could even use low petals. Role here have I used low petals rule. So you differentiate top and bottom with respect to end. So the numerator becomes foreign, the denominator becomes too and plus one. And here you'd still have infinity over infinity so you could use Low Patel once more low petals rule. So the numerator becomes just before the denominator becomes the two. That limit is two. However, that's bigger than one. So going on to the next page, we would be able to conclude that this series here is going to diverge. So let's let me write that out on the last page since the limit, How yeah, is bigger than one. The series okay, diverges by the ratio test, and that's our final answer.

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