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Center,2 |~2 -11072Find the radius of convergence and the interval of convergence of each of the series:(c) 2er n:...

Question

Center,2 |~2 -11072Find the radius of convergence and the interval of convergence of each of the series:(c) 2er n:

center ,2 | ~2 -1 10 72 Find the radius of convergence and the interval of convergence of each of the series: (c) 2er n:



Answers

Find the radius of convergence and interval of convergence of the series.

$ \sum_{n = 8}^{\infty} \frac {(x - 2)^n}{n^2 + 1} $

All right. This problem talks about radius is of convergence and intervals of convergence of a series. And the key to answering these questions is these two steps here in red first we need to determine when the limit of the end plus first some or the end plus first term over the ants term is less than one. That will give us our radius of convergence and then we need to check our endpoints individually to see if we can include them or not in our interval of convergence. Okay, so this example is going to have something a little different in it. We're gonna look at the Some from n equals 2 to infinity of and here's one of the big differences. It's not just an X to the end anymore. It's an X plus two to the end divided by two to the n natural log of n. Ok. So that X-plus 2 to the end. That is going to come into play when we find our interval of convergence and we'll talk about that a little later. But let's get started on finding our radius of convergence. We do that still the same way we're going to take a look at the limit as N goes to infinity of the end plus first term. So again, everywhere we see and and we're gonna put an end plus one. So we're gonna have an X plus two to the end plus one over To to the end plus one Times The Natural Log of and Plus one. So that's my numerator of my fraction and then divide by the denominator which is just the same stuff. But with ends instead of And plus once we're going to have X plus two to the end over to to the end natural log event. All right. So, we need to calculate that limit. Step one, Let's flip our fractions and multiply instead of dividing. So limit as N. Goes to infinity X-plus 2 to the end plus one Over 2 to the N Plus one. Natural log of n plus one times now by the to to the end. Natural log of end Over X-plus two to the end. Okay, So again just flipping and multiplying there. Now let's check what cancels out because things always cancel out in these problems. I have an X plus two to the end plus one, divided by an X plus two to the end. So those I can simplify and get just an X plus two. Okay, I have a two to the end and a two to the N plus one in the denominator. So that will leave me with a two in the denominator. Okay. And then the natural log of N. And the natural log event plus one. I can't combine those at all. Those aren't the way my log rules work. So those are just going to have to stay put. Okay. So when I simplify this, I'll get the limit as N. Goes to infinity Absolute value of X-plus two of virtue. And then times the natural log of N. Over the natural log of N Plus one. Yeah. Okay. Already great. Now that X plus two over to that doesn't have any ends in it. So it can factor out of the limit. So X. Plus to keep that in absolute values divided by two, Take the absolute value of two and we get to so X plus to an absolute values over two times the limit as N goes to infinity. Uh The natural log of N. Over the natural log of N Plus one. Okay. Alright. And that's all in absolute values. Now that limit is not one we've probably seen before. So we maybe don't know that it what it goes to. So let's take a look at it and see what we can get. Mhm. So as N goes to infinity, the natural log of N goes to infinity. And the natural log of N Plus one goes to infinity. So I have an indeterminant form here, infinity over infinity. Which means I should use low petals rule. Okay, so lumpy tiles rule says again, let me just put here. We're doing low petals rule. Low petals rule says. Okay. The limit is the same as if you took the derivative of the numerator. So the derivative of natural log of end Is one over end. And divide that by the derivative of the denominator, derivative of the denominator would be one over N plus one. Okay. And now life is better Absolute value of X-plus two over to on the outside limit and goes to infinity. Simplify those fractions and you should get something you recognize And plus one over N. Right. And that limit goes to one. Okay, so this limit is equal to the absolute value of X plus 2/2. Super. Okay. Now, according to # one up here, we need that limit to be less than one. So I need Absolute value of X-plus two over to To be less than one. So if I multiply both sides by two I get the absolute value of X-plus two Is less than two. Okay, that too is my radius of convergence and my radius of convergence is equal to two, awesome. Okay, now, so far we haven't had to worry at all that it's an X plus two instead of an X. That's going to come into play now when I try to find my interval of convergence. Okay. All right, so um for my interval of convergence yeah, I need absolute value of X plus two to be less than two, so that tells me. Okay, um I'm gonna look at an interval centered around negative too again because this is X minus negative too and I need to go out to in both directions. Alright, so that's going to get me to zero here and down here too negative for Okay, so my interval I know it works for all x between zero and 4 but I need to check out the endpoints of that interval. Can I include those in my interval of convergence? Okay. All right, so let's check those out. So when X is equal two negative four. Okay, again, we're going to just plug negative for into our initial problem here for X. Okay, So when X is -4, this is going to look like the some and equals 2 to infinity negative four plus two to the end. Mhm. Over to to the end times. Natural log of an. Okay, let's simplify that. See if that series converges and equals 2 to infinity -2 to the N Over 2 to the N Natural Log Event. Alright, my negative 2 to the end and the two to the end would get me a negative too over to to the end which will get me a negative one to the end divided by natural log event. Mhm From N. equals 2 to infinity. This converges by the alternating series. Test it alternates and the terms go to zero. Natural log of end goes to infinity, so one over it goes to zero. So again by the alternating serious test This series when X is negative, four is going to converge. So I can include that in my interval of convergence. All right. What about when X0. Yeah. All right. Let's check that out. Plug in X equals zero. So I'm gonna add up from two to infinity of zero plus 2 to the end Over 2 to the N Natural Log of N. All right. Let's do some simplifying and he goes to to infinity two to the N over two to the N. Natural law. Again, simplify awesome. The to to the ends are going to cancel each other out Some and equals 2 to infinity. One over natural log of n. Okay. Now, what about this series? Does it converge or does it diverge? Okay, well, let's take a look and let's compare it. Let's compare the natural log of end to end. We're going to do a comparison test when anna is greater than to the natural log event is less than M. So one over the natural log of N Is greater than one over end. Okay, so my series here is greater than the series From an equals 2 to infinity of one over n. This is my harmonic series. This diverges. So by the comparison test, anything bigger than it diverges. Mm. So therefore my original series also diverges. And so to finish out the problem here, my interval of convergence, mhm. As we talked about it was going to go from negative for 20 and I needed to check the endpoints and what I've learned by checking the end points as I can include The -4. That's here by the alternating series Test, but I cannot include the zero because then I'm bigger than the harmonic series and that already diverges. Okay, so again, this is what it would look like in um in inequality form. Or if you want interval form, it would be from negative four With a Bracket Square Bracket 2, 0 with a round parentheses. E to show we cannot include that. Alrighty. Great job. You're getting it. See you next time.

Okay for this problem we'LL use the ratio test first. Just figure out where we get convergence. So Lim has n goes to infinity of a and plus one over a n and I and we mean this whole chunk here, including the X values This is X to the two times in plus one over and plus one Sheldon of n Plus one squared. So that's just our A And plus one term we're dividing by a m which is multiplying by the reciprocal the van. So what now We're multiplying by n times natural log in squared and we're dividing by X to the two n so x to the two times implicit One is the same thing as X to the two n multiplied by X squared so extra too And we'Ll cancel out with this x to the two ends and we'Ll just have x squared there and then here we have an end here we have an n plus one Well, group those terms together and then here This is an exponent of two This is an exponent of two. Well, also lump these terms together as n goes to infinity Natural log of and divided by natural law. Govind plus one is going to go toe one. You can see that by applying low Patel's rule and as n goes to infinity and over in plus one is also going toe goto one. So this is just going to be absolutely You have X squared and we want for that to be less than one. Okay, but X squared is already going to be something that's non negative. So x squared. The only real condition here is that X squared is less than one. Okay. And this means that axe is going to be between minus one and one. So we get that by taking the square root of both sides square root, and then doing the plus minus. And you get that exit between minus squared of one and positive squared of one which is just minus one and one. Okay, so our interval of convergence is somewhere from minus one. No one. At this point, we don't know whether or not we include minus one and whether or not we include one, the length of this interval is going to be to go, which means that the radius of convergences too divided by two, which is one so one is our radius of convergence. To figure out the interval of convergence, we need to know what happened when we plug in minus one into here. And what happened when we played in one into here? Okay, but minus one to the two n is just going to be one, because the exponents going to be even and one to any power is going to be one. So we're going to get the same case when X is equal to minus one. That'LL be the same thing as when X is equal to one. So there's really only one case we need to check here. Okay, So when excited, too. And get your place with one. We need to ask ourselves what's happening then We have one over in time's natural log in squared, and here we can use the integral tests to figure out whether or not we get convergence. Okay, so figure out whether or not this integral is convergent or not. Case will be where change of variables here. So we'LL do u equals natural log of n which means that d'You is one over in d n A. So whenever we see a one over Indian and this equation will replace that buy, do you? Okay. So if n is equal to two, then you is equal to natural log of two. When n is equal to infinity, you is equal to natural log of infinity, which is still infinity. Here we have a one over and Deon So we replace that buy, do you? And then here we have natural log of end but natural lot of vintage issue. So here we just have one over you squared one over you squared is just you to the minus. Two of you write it like that. It might be more clear how you evaluate this. We just do the power rule for inter girls. So this is going to turn out to be minus one over you evaluated from natural log of two up to infinity. Can we like to think about this is a limit, So All right, Lim eyes, We use him. Hear his m goes to infinity minus one over you from natural log of two, two em. So that's limit. His m goes to infinity one over. Sorry, minus one over. M minus minus one over. Natural log of two. This term is going to go to zero. Here we have negative. Negative. So that's going to turn out to be something positive. But the important part here is that this is this is something that's finite. So the interval is going to converge. So by the interval test, we get that the some that were considering is also going to converge. Okay, so both of these in points are going to be included. So our interval of convergence, we include minus one, and we include one.

To figure out the radius of convergence will use the ratio test here to take the limit as n goes to infinity, absolute value of and plus one over and and by and we mean this whole thing. So including the X values here. So this is limit as n goes to infinity of absolute value of X plus two to the end, plus one over two to the end, plus one natural log of in plus one to his are and plus one term divided by a ends multiplying by the reciprocal. We're multiplying by two in natural lot of n divided by tax plus two to the end. So X plus two to the n plus one divided expert to to the end that she's going to leave us with X plus two two to the end of two by two to one plus one is going to leave us with one half. And then we have Ellen of in divided by Ellen of and Plus One, and to figure out the limit as n goes to infinity of natural log of n divided by natural log of one plus one. You do low Potala jewel here so both the top and the bottom blow up to infinity. So you, Khun, apply low Patel's rule Look, tiles rule. You do the derivative of the top divide. By the derivative of the bottoms, we get one over in divided by one over in plus one. So we'd get limit as n goes to infinity in plus one over N, which is just one. So Ellen of end over Alan ofhim plus one does goto one. So this is just absolute value of X plus two. Oh, her too. And we want for that to be less than one. So if we multiply both sides by two, we get absolute value of X. Plus two is less than two. So at this point, you might already be able to see that the radius of convergences too. If you can't, the way you figure it out is if this happens, then X plus two is trapped between minus two and positive too. It's of minus two is less than X plus two. If we subtract two from both sides, that means that minus four is less than X. And if we have X plus, two is less than two. If we subtract two from both sides. We get that X is less than zero. So the length of our interval of convergence is zero minus minus four. So the length of our interval is for the radius of convergence of half of the length of the interval of convergence. So half of four is two. So that's that's our radius of convergence are for the interval of convergence. We need to figure out whether or not we want to include the end points. So whether or not we include X equals zero and whether or not we include X equals minus four. Okay, So if X is equal to zero, then we'd have to do the end over to to the end. Sorry, the two to the end Over to land will cancel. Just have one over natural log of end came in. If N is equal to two or any larger than two, then natural lot of N is going to be something that's smaller than just regular. And if we replaced the denominator was something that's bigger than we should get, something that's even smaller. Okay, so this should be larger in this sum and this sum close up to infinity. That means that this some also has toe blow up to infinity. So this is divergent. Okay? And then we need to figure out what happens when X is minus four. When X is minus four, we'd have minus four plus two, so we'd have minus two to the end. So the two to the ends. Well again. Cancel. Then we have minus one to the end. Happening when X is minus four. Plugin minus four under here. What we get is the sum from n equals two to infinity of minus one to the end. Over natural log of n. And this is going to converge by the alternating signed test. So we do want to include minus four, but we do not want include zero zero gives us divergence minus four. Does not so minus for we include that zero is bad. So we throw that out


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