[IOpts] Consider the series: 46-2What is the cerler of the series?b) Find the interval of convergence of the series and determine the radius of convergence R...

(a) What is the radius of convergence of a power series?
How do you find it?
(b) What is the interval of convergence of a power series?
How do you find it?

Okay for this problem, We're going to deal with some concepts. Here are a So we're going to first determine what's the radius of convergence to remember that for a power seriously, in the form that it's expanded at a point as it was to a would always have a, uh, they constrained that this is only valid inside the neighborhood, which is X minus a, um, the absolute value of accidents has to less than are here. This are we call it the radius of convergence. And for part B on how do we find the greediest convergence? So we're gonna first find this relationship where the distance between accent is less than our So we call it are the radius of convergence. So let's say if we know that we have the interview of convergence is going from Let's see t and s. So the rays convergence this city and this is s This is X. So let's say as it's created anti without the loss of generality or is gonna be equals to the distance between as empty, divided by two. Okay,

In this problem, we have been given a series and we need to determine the radius of convergence of it. Now. The given series of the problem is x minus x square by two plus X cubed by three and so on. So that the NFL film is minus one to the power, n minus one. Actually power and divided by hand and the series goes on. So from here we can see that this series is equal to submission and is equal to one to infinity minus one to the power and minus one, divided by end times X to the power. And from here we can obtain that. The innit coefficient C n is equal to minus one to the power and minus one by end. And if we replace N by N plus one we will obtain C N plus one is equal to minus one to the power and plus one minus one. So that's just and divided by n plus one in order to determine the radius of convergence. First of all, we need to determine the value of key, which is equal to the limit as N tends to infinity. The modelers of C N plus one divided by the modelers of CNN. So that will be equal to the limit as intends to infinity. The model is of cm plus one will be one divided by N plus one minus one. To the power in turn gets removed because of the absolute value sign and in is a positive number because it's a natural number. Seven plus one would be positive and this one by endless one would be positive. Similarly, the models of CN will be equal to one divided by end. So this will be equal to the limit as intends to infinity of N divided by N plus one. And in order to determine this limit, let us divide the numerator and the denominator by n looking one divided by one plus one by 10. Now, one day in tends to zero as intense to infinity. So this will be one by one plus zero, which will be one very by one, which is equal to one. So the radius of convergence are will be equal to one by key, which is one by one. So this will be equal to one and the radius of convergence is one. Now we have also been asked what this tells us about the interval of convergence of the series. Now, not that series. This series is Britain in terms of X to the power in which is X 0 to the parent. So it is centered at the origin zero. Hence the radius of convergence being one. If the interval of convergence of this series is I Then we will have the open interval -101 as a subset of the interval of convergence. I now in the next problem we need to investigate the convergence at the end points of the interval of convergence. So if we consider the point minus one, then the series becomes X. So minus one minus X squared by two, su minus one square, which is one divided by two plus plus X cubed by three, so that's minus one cube, so minus one, provided by three with a minus one by four and so on. So this is the negative of the harmonic series, one plus a half plus one by three and so on. So this will devote, so that means the series diverges at the point X is equal to minus one. Now if we consider the point X equals to one, we have one minus x square by two, so one square by two plus X cubed by threes, one divided by three minus one, divided by four and so on. And by the alternating series test this will converge because the terms are of the form one by N and so A N plus one will be less than a N. And also the innit term will tend to zero as intends to infinity. And by the ordinating series test this will convince. So that means that the Power series does not converge for the 0.-1, but it converges for the .1. So this will be the intervals of convergence -1 less, the next battery close to one.

Okay, so for this Siri's that be to the times X minus in expense A. To the entire world. The National League win or installed from two? Because, interestingly, at least one here we have this part goes to zeros, which is invalid. Um, we have a lot of terms within so in tears on power term. And it's also in the function term, we're gonna do a racial test, which will be very convenient. So the region has reads the limit of the absolute value of a consecutive two terms has to be less than one. So this is going to give us the absolute value of X months age less than one will be hence R equals to one it would be. And right now we have X s from a minus. One would be to a plus, one will be that's considered the quarter cases. So if X equals to that one that's called X left. This is X right. So X equals two. Expert. What's gonna made a serious though? So that's going to go to a like you want to the harbor or natural log in, and we know that this is convergence. Siri's by the ordinating serious test and, uh, on the other corner we have when over National Live in. And this is going to be a divergence. Siri's, since this is even greater than another deputy, Siri's very well known, which is the harmonic series so we can conclude that the interval of convergence will be a minus. Bi, uh, minus one will be to a plus. One will be, but the left hand side is included.

Hey, to figure out the radius of convergence, we do limit as n goes to infinity of absolute value of a N over absolute value of and plus ones, absolutely of this whole thing where the ends air just one over into the fourth times for the end. This is limit as n goes to infinity Ah, in plus one to the fourth times four to the end, plus one divided by into the fourth times for it to the end. So this's limit as n goes to infinity of thin plus one over in to the fourth and then for the n plus one divided by four to the end Oops, sorry. I forget that for the M plus one divided by four the end that's going to cancel out just before and then and plus one over in as n goes to infinity and plus one over n goes toe wants This is one to the fourth, which is just one. So one times for so just get for here. So four is the radius of convergence Here for the interval of convergence, we need to check whether or not four works when we plug it in here and whether or not minus four works when we plug it in here. So when axes equal to minus four, then we have minus four to the end over into the fourth times for the end. Sorry for the bad hand writing. Okay, so minus four to the end, divided by four to the end, that's going to turn into a minus one to the end because minus forty and is minus one of the end times for the end and then four to the and will cancel out before the end that we have down there. Okay? And then this will converge, but the alternating signed test. And if we look at, if we look to see what happens when X is equal to four, then we would end up with for the end divided by four to the end. So those would cancel. And we just have one over and to the fourth. Since four is a number that is larger than one, this would be something convergent. So this would also converge. Okay, that works. So minus one works are starting minus four works. When you plug it in for works, when we plug it in. So We include both minus four and foreign R Interval of convergence. So we use these close brackets. Tau show that we are including them and that's gonna be our answer.