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Test the series for convergence or divergence. $…

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Problem 10 Medium Difficulty

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

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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 this series for convergence or divergence. Now the first thing I notice is that this series is alternating. That's because due to this negative one to the Empower so this suggests that we try the alternating series test. Okay, so here we define BN to be squared of end over two n Plus three. This is positive because numerator and denominator are both positive. So that's the first condition that has to be checked. The second condition. We need the limit as n goes to infinity of bm to be zero and this problem that becomes end to the one half over two n plus three and this limit is zero. And if you can't see why, you can use low Patel's role here, mhm and then finally we have one more condition to check. We need that BN is decreasing. So when we add one to the end, it's not bigger than the previous value of being. So to check this, we could go ahead and actually plug in and plus one into our formula. But as mentioned in the text book, another way to show that bien is decreasing if we have f prime of X is negative. Where we define f of X to be X likes a squirt of X over two X plus three and then we can write. This is X to the one half over two X plus three. So now if we can show that this function f has negative derivative, that's equivalent to showing that this function is decreasing and that will complete the last condition here. And then finally, we would be able to say that the series converges by a S T alternating serious test. So let's go to the next page and show that F has negative derivative. So this is X to the one half two X plus three. So f prime of X is the quotient rule. So the denominator is positive because it's a square. So what we're really interested in here is the numerator. Yeah, so here we can go ahead and write this Mhm. Okay. All I did was just distribute this term out here inside the parentheses and then simplified. Now this is equal to three. Now, let me go ahead and combine these two so that will be minus X to the one half over the square and then get a common denominator, and we see that this will be negative, if so, three minus two x so that that previous term was negative. If this is less than zero, so three is less than two works. 3/2 is less than X. So this tells us that bien is decreasing when and is bigger than or equal to two. And that's what allows us to use the alternative series test. So all conditions okay for BN in the alternating series test. MM hold. Therefore, the series converges by the alternating series test, and that's our final answer.

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