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

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

$$I=[1,3]$$

Calculus 2 / BC

Chapter 11

Infinite Sequences and Series

Section 8

Power Series

Sequences

Series

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okay for this problem will use the ratio test toe figure out when we get convergence here for the ratio test we limit is n goes to infinity of a n sorry and plus one over and with a in terms is this whole thing So including the X values here. So that's limit as n goes to infinity of absolute value of X minus two to the n plus one divided by in plus one squared plus one. So that's our and plus one. And we're dividing that by a m so multiplying by the reciprocal. So we're multiplying by n squared plus one over X minus two to the end. Okay, so X minus two to the M plus one divided by X minus two to the end is just going to be X minus two and then we have in squared plus one divided by in plus one squared plus one. So these up top and down below we have degree to polynomial is with leading coefficient one. So as n goes to infinity, this is just going to go toe one. So this is just going to turn out to be absolute value of X minus two, and we want for that to be less than one. Okay, so absolutely of X minus two is going to be less than one when X minus two is trapped between minus one and one. So if we have that minus one is less than X minus two. We can add to to both sides and see that that corresponds to one is less than X And if we have X minus two is less than one we can add to to both sides to see that that corresponds to X is less than three. Okay, so the interval were working with is from one to three. We're not sure whether or not we include one and whether or not we include three at this point. But the length of the interval is three minus one, which is too to the radius of convergence is half of that to the radius of convergence. R is half of the interval of convergence half of the length. The radius of convergence is just one, and we need to figure out whether or not we include X equals one and whether or not we include X equals three for the interval of convergence. So in X is equal to one. If we plug in one here we have minus one to the end, divided by n squared plus one. And this is going to converge by the alternating signed test. That's good. And the other thing we're wondering about is X equals three. So if you plug in X equals three here, then we get three minus two to the end. So that's one to the end. One to the end is just one. So we would get some just won over and squared, plus one, one over in squared plus one is goingto converge because one over in squared converges the exploding here is something that's bigger than two. So we will get convergence here and here. The denominators even bigger, right? If we make the denominator bigger, we're gonna get something smaller. So since this converges, we have to have that this converges. So three works. One works. So we include both three and one in our interval of convergence

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