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(a) Show that for each integer n 2 1,+nn(n + 1)Use part (a) to show that for each integer n 2 1,n(n+4)This justifies one of the summation formulas we used earlier i...

Question

(a) Show that for each integer n 2 1,+nn(n + 1)Use part (a) to show that for each integer n 2 1,n(n+4)This justifies one of the summation formulas we used earlier in the course:Determine whether the seriesIS convergent

(a) Show that for each integer n 2 1, +n n(n + 1) Use part (a) to show that for each integer n 2 1, n(n+4) This justifies one of the summation formulas we used earlier in the course: Determine whether the series IS convergent



Answers

Determine whether the series is absolutely convergent. $$\sum_{n=1}^{\infty} \frac{(-2)^{n} n !}{(2 n) !}$$

If this series is absolutely convergent here. So basically then, well, we Not from the behavior of our 10 that are tangent approaches fly over to as n goes to infinity here. And so this is always positive up here. Yes, we're taking the value of this. Um then we're considering this negative one to the end, go away. So let's take a look at that. Stop being able to arc 10, pop and over and squared. And so then we have here, since our 10 event is always going to is going to approach the pi over two, then we know that that's going to be less than or equal to Pi over two over and squared. Then from here, you know that this coverage is by P series. Therefore this converges by comparison to the P series and so therefore this the series is absolutely convergent.

This is absolutely conversion here. There is no negative ones at the end here. So we'll just take the series as it is here and let's apply the ratio, test and take the limit as N approaches infinity of N plus one. Said the second power over To to the end plus one. We're dividing that by and squared over two to the end. Simplifying this down fully here or simplifying it by multiplying by the reciprocal to the bottom there. We have the limit as an a purchase affinity of N plus one over to the end plus one Times 2 to the end over and squared. Yeah. To discredit him. So and this goes to infinity this these both go to once. It's their highest power are and squared for both. And the coefficients are number one and then two hints cancel that they used to in the bottom there. So that means that the evaluation to limit is equal to one half. She's definitely less than one. So then we know that this is absolutely convergent. So the serious is so

Check and see if this series is absolutely convergent, Which means that we have to take the opposite value 1st. So we take the opposite value of this serious. We are left with Squared event over and plus one here. Okay, So then if we take that as and approaches infinity here and we see actually, let's just compare this with sees the comparison test and if we multiply Squared event and plus one times one over square event to top to bottom, then we're left with one over squirt of N plus one over a radical hand here, which means that this is actually going to be less and see here. So okay, so actually speak this this way. So if we that would be greater than one over to radical in here and we know that. So for the P series this diverges P serious, so therefore this would also the average. But the comparison tests to the P series that we have there. And so then this series is not absolutely convergent.

So because no submission of the seas convergent so from enough. Yeah, natural enough am and must be smaller than one in actual beacon virgins. And that's why we will have Yeah, I and it would be 1,000,000 square will be smarter than I am. And, uh, and then because of submission that mean is in virgin Didn't buy comparison. There's can conclude that s immersion in square on so convergent


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