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~12 points SCalcET8 11.8.006.Find the radius of convergence, R, of the series.(~1Yx n9 n = 1R =Find the interval, I, of convergence of the series. (Enter your ans...

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~12 points SCalcET8 11.8.006.Find the radius of convergence, R, of the series.(~1Yx n9 n = 1R =Find the interval, I, of convergence of the series. (Enter your answer using interval notaNeed Help?Rcad ItTalkto Tutor~12 points SCalcET8 11.8.007.MI.Find the radius of convergence, R, of the seriesxn + 2 Znl = 2Find the interval, I, of convergence of the series: (Enter your answer using interval notaNeed Help?Read ItWatch ItMaster ItTalk to Tutor

~12 points SCalcET8 11.8.006. Find the radius of convergence, R, of the series. (~1Yx n9 n = 1 R = Find the interval, I, of convergence of the series. (Enter your answer using interval nota Need Help? Rcad It Talkto Tutor ~12 points SCalcET8 11.8.007.MI. Find the radius of convergence, R, of the series xn + 2 Znl = 2 Find the interval, I, of convergence of the series: (Enter your answer using interval nota Need Help? Read It Watch It Master It Talk to Tutor



Answers

Find the radius of convergence and interval of convergence of the series.
$$\sum_{n=2}^{\infty} \frac{(x+2)^{n}}{2^{n} \ln n}$$

But the submission off the minus one bulb and expert and even by far about in terms and and from Judge Infinity in this question Here, we need you a blind, uh, racial past here. So we have a racial test, and then we need to compute the limit under I am this one over. I am and goes to infinity. And there was we get now the limit off now for, uh, I am this one which again, the exponent m this one off the farmer and this one times and then I'm the and bless one. And now we divide by and so we should. And you're multiplying by the reciprocal. So should get now will be farmer and times and in the end, dividing by Expo. And here we see, we can consider the expert. And with this power Farber and with this power here and then we can bring the X and a form here outside a limit, if always again. No ego do the absolute X over four times the limit off and and over the Ellen off em, that's one. And as we see that as an goes to infinity, this limit here where you could you one Therefore, we should get a coaching option of angst over far. And now, Judy, Convergence, This limit here must be smaller than one on doesn't even only if absolutely, banks with monitor and car, even only if absolutely, x will be actual baby Dream man is fine for here and now for the first case for the left and point him, Mexico Germany's far We get a serious now will become the submission, uh, minus $1 minus far about. And they're ready for about and and from Judge Infinity. So we cancel out everything and then we have a submission every one of us and then in here. And we know that this one will be divergent because by the Comm Person Tasmania one of an enema and is bigger than one of n. And we know that there's so much in under one of Andy's diversion by the harmonic series here. Now find the exit good. You far. We have the X go too far, and then we will have the service that will become the submission contusion, infinity. And then we have ah minus one about under and invaluable and and and this is his alternating Siri's. So that's where we diverge in by, uh, the meat off the one of l a and and goes to infinity Koji zero. And then we can go that the interval we have here it will be from one is far too far. And because we have the divergent on the ankle bituminous far. So you just up in interval and that region in the expletive five square. So we just uh huh. Now, this will be sorry. It doesn't beacon vision here. Sorry about that. Isn't the convergent. And they're far for this one was again the close in the vote here and now And for the radius here And where you go, Joe Far.

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.

For the radius of convergence. We take the limit as n goes to infinity Absolute value of a n over a n plus one. And we want for this to be less than one. The ends here is this to to the end times and squared. Okay, so up top, we're gonna have to the end and squared down in the denominator We have to do the in plus one times in plus one squared. So to the end about about to the M plus one is one half and then n squared over in plus one square it is, and over in plus one squared limit as n goes to infinity of n over in plus one is one this turns into one squared so one half times one squared would just get one half So one half is the radius of convergence. So this is our are for the interval of convergence. We need to figure out whether or not we include minus one half and whether or not we include one half case, we need to check to see whether or not we get convergence for those values. Okay, so this was the sum that we had we had to the end. I'm in squared times X to the end, and we were going from an equals one to infinity. So one two check are in points. So we needed the check Well, minus one half. So plug in minus one half here and see whether or not we get convergence. So when you plug in minus one half for acts, we get minus one half to the end. So if we rewrite that a little bit, we get minus one to the end times one half to the end. So to the end, one half to the end cancel. And this should be clear that this is not going to converge because the terms do not go to zero. So that is going to diverge. And similarly, when exes one half the terms that we get are going to be two to the end and squared one half to the end. So the terms that were going to be summing up is just in squared, which clearly do not go to zero. So we can't possibly have convergence, So that's also went to diverge. So both of the end points we have to toss out so the interval of convergence. We throw out both minus one half and one half and just leave it as this open interval here.

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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