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

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

$$[-4,0)$$

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Oregon State University

University of Michigan - Ann Arbor

University of Nottingham

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