Like

Report

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

$ \sum_{n = 1}^{\infty} \frac {5x - 4)^n}{n^3} $

$$I=\left[\frac{3}{5}, 1\right]$$

You must be signed in to discuss.

Missouri State University

Campbell University

Baylor University

University of Michigan - Ann Arbor

okay for this problem, I'm sure they're supposed to be Ahh, princesses here. So it should be quantity five X minus four to the Andover and cubed toe throughout the radius of convergence Weaken, Use the ratio test here. Tto figure out when we get convergence So that a n here is this whole chunk, including the X values. So this is goingto tell us when we get convergence and this is going to simplify to five X minus four times and cubed, divided by in plus one cube. And when we're doing the ratio test, we want for this to be something that's less than one. Since these have the same exponents, we can combine them like this, as in goes to infinity and over in plus one goes toe one. So this is going to turn out to be one cube, which is just one. So this is just five x minus four. And when we're looking to see where we get convergence again, we want for this to be less than one, which means five X minus four is trapped between minus one and positive one. We can add four to both sides and appear. Then we can divide both sides by five, and that's not goingto flip the inequality signs here. Write. This is still less than it's still less than it's not going to flip the inequality sign because we're divided by five and five is a positive number. We're dividing by a negative number. Then they would have toe flip. Okay, so the interval of convergences goingto be from three over five toe one can. At this point, we're not yet sure whether or not we include through over five and whether or not we include one. So the length of this inner blues one minus three over five to the radius of convergence is the length of the interval divided by two. Okay, so we just do some algebra here. One is just one over one. So get a common denominator between these fractions. So this is five minus three over five, divided by two. So that's five minus three is too. So this top part turns out to be two over five, and then we're dividing by two, which is the same thing. It was multiplying by the reciprocal. So they're typical of two is one half these twos cancel. We'LL get that the radius of convergence is one fifth. Okay, Now we need to figure out for the interval of convergence. Do we include three fifths? Do we include one? So if X is three fifths, what happens here? So if X is three fifth, then we have five times three over five minus four. So five times three over five is three. So this is going to be three minus for which is minus ones. We have minus one to the end, and we're dividing by N cubed. So this we're going to converge by the alternating signed test. And then the other value we need to check is X equal, the one So if X is equal to one, then we have five times one minus four. So five minus forward to positive one one to the end. It's still one. And this is still going to convert because the exponent here's three. Three is bigger than one. If the exponents was one or anything smaller than we would have divergence. But since three is bigger than one, we get convergence. So we include three fifths, and we include one. So that's our interval of convergence