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Test the series for convergence or divergence. $…

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Problem 11 Medium Difficulty

Test the series for convergence or divergence.
$ \displaystyle \sum_{n = 1}^{\infty} (-1)^{n+1} \frac {n^2}{n^3 + 4} $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 5

Alternating Series

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Video Transcript

let's test the series for convergence or divergence. Now the first thing that I see is that this series is alternating. I know this because of the negative one to the end, plus one power. So this suggests we could try to use alternating series test. So to do so, we should define Bien to be the positive part. So we ignore the minus one and just write down everything else. And then there are three conditions. The first one is that BN has to be positive check. That's true because numerator and denominator are both positive. Then the second one we need that the limit of B n A zero. And this is also true. We have n squared over n Q plus four if you want to show this one way to do it, for example, is to just use Low Patel's rule. Okay, and there's one more condition. Yeah, we need that the sequence be in is decreasing. So one way to show this is to show by hand that BN plus one is less than or equal to be in. And the difficulty of doing this really depends on the example for this problem. I would not really want to use this because if I replace end with and plus one, then when I plug this in to be in, I'll have to also square in Cube, the N Plus one. And I prefer not to do that if I don't have to. So the other method shown in the book is to define f of X by just replacing and with X in the formula for BN. The reason for doing this is because now we're we're just extending the function b from these values of N, whereas now over here X could be any real number. So now we can take the derivative and the question is whether the derivative is negative, because if this is true, this is equivalent to saying that F is decreasing. Recall. The interpretation of the derivative is the slope. But if F is decreasing, that's equivalent to be and decreasing because F is just to find in terms of B b n. So now we'll go to the next page. We'll find this derivative and the hope is that it is eventually negative. And then we could say that the series converges by all streaming series test so we'll go on now to the next page. Uh huh. This was F. So let's compete that derivative. Here you can use the quotient rule so derivative of X squared. And then I take that denominator and then I leave X squared. And then I do derivative of the denominator over the denominator squared. Now the denominator is not a problem for us because it's positive. So the question is whether the numerator is negative. Mhm. So let's simplify this. So I have. This will be two x to the fourth plus eight x minus three x to the fourth. Keep that denominator as it is now in the numerator. I can simplify this by pulling out of X and then I just come. I combine these together to get a negative X to the fourth. And after pulling out of X, I just have X cubed. And now this thing will be negative if eight is less than X cubed. So it's also, in other words, X is just bigger than two. This tells us that bn the sequence bien starts decreasing when n is bigger than or equal to three. So let's just say it starts decreasing when n is three. So this means b n plus one less than or equal to B N. If N is three or more and therefore we have satisfied the third condition for the alternative theories test that the sequence bien is decreasing. Yeah, therefore, going on to summarize the series converges by the alternating series test, and that's our final answer.

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