Radius and interval of convergence Use the Ratio Test or the R...

Question

Radius and interval of convergence Use the Ratio Test or the Root Test to determine the radius of convergence of the following power series. Test the endpoints to determine the interval of convergence, when appropriate.

x +x³/3 +x⁵/5 +x⁷/7 + ...

x +x³/3 +x⁵/5 +x⁷/7 + ...

Accepted Answer

Hi, everyone. Let's take a look at this practice problem. This problem says to find the radius and the interval of convergence for the power series of Y divided by 2, plus Y cube divided by 4, plus Y 5th divided by 6, plus Y 7th divided by 8, and so forth, using the ratio test. So here we need to find the radius and interval of convergence for our power series using the ratio test. Now we call for the ratio test that if we have a series that's of the form, that is the sum of a subn, then we need to find the limit L, which is equal to the limit, as N approaches infinity of the absolute value of the quantity of a n + 1 divided by a quantity. So we first need to determine what A of N is. Now, if we look at the power series that we're given, we notice that the exponent on Y and the numerator is always an odd value, and the denominator of the fraction is always an even value. So that means that a subn is going to be equal toy raised to the quantity of 2N plus 1 in quantity, and this is going to be divided by the quantity of 2N plus 2 in quantity. Wherein goes from 0 up to infinity. So that means that L is going to be equal to the limit as N approaches infinity of the absolute value of the quantity of the quantity of Y raised to the quantity of 2 multiplied by the quantity of N + 1 in quantity plus 1 in quantity, divided by the quantity of 2 multiplied by the quantity of N + 1. In quantity plus 2 in quantity and all that is divided by The quantity of Y raised to the quantity of 2N plus 1 in quantity, divided by the quantity of 2N plus 2 in quantity. And we can simplify this this expression. So this is going to be equal to the limit, as N approaches infinity of the absolute value of the quantity. Of Y2 multiplied by the quantity of 2N plus 2 in quantity, divided by the quantity of 2N plus 4 in quantity. And we can simplify this further by dividing both the numerator and denominator by 2. In doing so, this is going to be equal to the limit, as N approaches infinity of the absolute value. Of the quantity of the quantity of Y squared multiplied by the quantity of N + 1 in quantity, and that's going to be divided by the quantity of N + 2 in quantity. And so now we take the limit as n approaches infinity. Now, as n approaches infinity, the quantity of n + 1 in quantity divided by the quantity of n + 2 in quantity approaches 1. So taking this limit, we see that this is going to be equal to the absolute value of Y squared. And here we call that for our series to converge that L has to be less than 1. So that means that the absolute value of Y2d has to be less than 1. Taking the square root of both sides, we see that the absolute value of Y has to be less than 1. This means that our radius of convergence R is going to be equal to 1. So that is the answer to the first part of this problem. Now, for the second part, we need to find our interval of convergence. Now, since the absolute value of Y has to be less than 1, that means that minus 1 has to be less than Y, which has to be less than 1. So we want to do now is to check the end points of this interval to see if our series converges at these endpoints. So, we're look at the case when Y is equal to -1. In this case, we see that our series becomes the sum. Of N equal to 0 to infinity. Of the quantity of 1 in quantity, raised to the quantity of 2N plus 1 in quantity, and this is going to be divided by the quantity of 2N plus 2 in quantity. However, 2 N + 1 is always going to be an odd value. So we have -1 raised to odd power, which is always going to be equal to -1. So that means we can simplify this expression as being equal to. The sum Of N equal to 0 to infinity of minus 1 divided by the quantity of 2N plus 2 in quantity. So we can actually simplify this expression even further by factoring out a factor of 2 out of the denominator. And so, this is going to be equal to minus 1/2 multiplied by the sum of N equal to 0 to infinity of 1 divided by the quantity of N + 1. Now what we can do is shift our index from starting at N equal to 0 to N equal to 1. In doing so, we'll see that this sum is going to be equal to minus 12 multiplied by the sum of N equal to 1 to infinity of 1 divided by n. Now this here is a P series with P equal to 1. And since P is equal to 1, we know that this series is going to diverge. So that means that we do not include. The Y equals to -1 endpoint in our interval of convergence. So next, we'll check our other endpoint, which is going to be Y as equal to 1. And so here we're going to have the sum. Of N equal to 0 to infinity of 1 raised to the quantity of 2N plus 1 in quantity, divided by the quantity of 2N plus 2 in quantity. And so here we raise 1 to any power, we're always going to get back 1, so this is just going to be equal to the sum. Of N equal to 0, to infinity of 1 divided by the quantity of 2n plus 2n quantity, and again, we can factor out of 2 out of the denominator. So that means that this is going to be equal to 1/2 multiplied by the sum. Of N equal to 0 to infinity of 1 divided by the quantity of n + 1 in quantity and again we can shift our index to start at N equals 1. In doing so this is going to be equal to 1/2 multiplied by the sum of N equal to 1 to infinity of 1 divided by n. And again we have a P series here with P equal to 1. And since P is equal to 1, that means that this series also diverges. So we're not going to include this endpoint in our interval of convergence. So that means that our interval of convergence is going to be the open interval from -1 to 1. So that is the answer that we're looking for for the second part of this problem. So just to recap, we found that our radius of convergence R is equal to 1, and we found that our interval of convergence was the open interval from -1 to 1. So those are the answers that we're looking for for this problem. So I hope that this explanation has been useful, and I'll see you in the next video.

Radius and interval of convergence Use the Ratio Test or the Root Test to determine the radius of convergence of the following power series. Test the endpoints to determine the interval of convergence, when appropriate.

x +x³/3 +x⁵/5 +x⁷/7 + ...

Key Concepts

Radius of Convergence

Ratio and Root Tests

Testing Endpoints for Interval of Convergence

Watch next

Radius and interval of convergence Use the Ratio Test or the Root Test to determine the radius of convergence of the following power series. Test the endpoints to determine the interval of convergence, when appropriate.x +x³/3 +x⁵/5 +x⁷/7 + ...

Key Concepts

Radius of Convergence

Ratio and Root Tests

Testing Endpoints for Interval of Convergence

Watch next

Radius and interval of convergence Use the Ratio Test or the Root Test to determine the radius of convergence of the following power series. Test the endpoints to determine the interval of convergence, when appropriate.

x +x³/3 +x⁵/5 +x⁷/7 + ...