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Problem 26 Medium Difficulty

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

$ \sum_{n = 2}^{\infty} \frac {x^{2n}}{n(\ln n)^2} $


Interval of convergence is $[-1,1]$
Radius of Convergence is 1


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Video Transcript

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.

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