Radius and interval of convergence Determine the radius and in...

Question

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.

∑ₖ₌₀∞ (2x)ᵏ

Accepted Answer

Hello. In this video, we are going to be finding the radius and the interval of convergence for the given power series. The power series given to us is the infinite series starting at 0 and going to infinity for the function 3Y raise the power of N. Now, in this case here, because we have the function 3 Y and is raised to a singular power of N, we can go ahead and reduce this function, or evaluate it by using the root test. The root test asks us to take the limit as N approaches infinity of the nth root of the absolute value of A N. And this is where A N is the original argument given to us in our series. So by using the root test, we are going to represent the series as the limit, as N approaches infinity of the nth root of the absolute value of 3Y raise the power of N. This is going to further simplify to be the limit as N approaches infinity of the absolute value of 3Y. And this is going to simplify to just be the absolute value of 3Y. Now, in order for the root test to be convergent, the limit has to be less than one. And with that being said, in order for the quantity 3Y to be convergent, the absolute value of 3Y must be less than 1. Now, if we were to solve for Y and divide both sides of this equation by 3, we will get that the absolute value of Y is less than 1/3. And so what this 1/3 represents is this is going to be the radius of convergence for our given power series. So we found the first solution where the radius of convergence is going to be 1/3. Now what we need to do is we need to go ahead and identify the interval of convergence. Now from this statement, the absolute value of Y less than 1/3, we can go ahead and expand this, and this will give us that negative or Y must be between positive and negative 1/3. Now, what we need to do though, is we need to check the end points for this interval of convergence. In order to check the endpoints, we're going to take each value of the endpoint, plug it into our original series, and evaluate the remaining series. So starting with the left endpoint, when Y is equal to -1/3, plugging this in into the given power series is going to give us N starting at 0 and going to infinity. Of the quantity 3 multiplied by 1/3, raise to the power of N. This is going to further simplify to give us the infinite series of -1 raise the power of N. Now, the problem here is that this is an oscillating term. Since this is an oscillating term, this is not going to converge to any finite value. Therefore, the value of Y equal to -1/3 is going to be divergent, and it will not be included in our interval of convergence. Now we need to check the second endpoint where Y is equal to 1/3. When Y is equal to 1/3, we will be left with 3 multiplied by 1/3 raise the power of n. This will further simplify to leave us with one race the power of n which also just simplifies to be 1. But this is also going to be divergent as well. And so what this means is that at both ends of the interval, the both endpoints are going to be divergent and not included in the interval of convergence. So with that being said, our interval of convergence is going to be from -1/3 to 13, not including the endpoints. And these are going to be the solution to this problem, with the overall solution being a. So I hope this video helps you in understanding how to find the radius and the interval of convergence, and we will go ahead and see you all in the next video.

Radius and interval of convergence Determine the radius and interval of convergence of the following power series.

∑ₖ₌₀∞ (2x)ᵏ

Key Concepts

Power Series

Radius of Convergence

Interval of Convergence

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