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₌₁∞ (kx)ᵏ

Accepted Answer

Hello. In this video, we are going to be finding the radius of convergence and the interval of convergence for the given power series. Now the power series given to us is the infinite series starting at 1, going to infinity of N2d multiplied by Y, raise the power of N. Now, by taking a look at this given series, we can go ahead and approach this by using the ratio test. The ratio test states the following. We can represent the infinite series as N approaches infinity of the absolute value of A N + 1 divided by AN. So what this means is that in order to find the radius of convergence, we are going to take the infinite limit of an absolute value of a ratio. In the numerator, we are going to have our argument with N + 1 plugged into it, and we are going to divide that by our original argument. So, in order to set up the ratio test, we will have the limit as N approaches infinity of the absolute value of our series with M+1 plugged into it. So that is going to be the quantity M + 1 squared multiplied by Y, raise the power of N + 1. And this will be divided by the original series argument, which is N2d multiplied by Y, all raise to the power of N. Now, what we want to go ahead and do is we want to simplify the limits. So, the first thing we can go ahead and do is distribute that exponent of M + 1 to every term in the numerator, and distribute the N exponent to every term in the denominator. That is going to leave us with the quantity m + 1. Raise the power of 2 multiplied by M + 1. Multiply by Y raise the power of N + 1, and this will all be divided by N raise the power of 2n, multiplied by Y raise the power of N. In the numerator, the exporting 2 multiplied by N + 1 will simplify to be 2 N + 2. And so now what we want to go ahead and do is we want to go ahead and represent this as a product of terms. So we can create a product of two ratios where we have a product of N terms and a product of Y terms. Now, let's go ahead and simplify each of these independently. For the first term, because both of these quantities are raised to the two N power, we can rewrite this first term to be N + 1 divided by N, raise the power of 2N. In the numerator for this 2 n + 2 this will separate to be m + 1 raise the power of 2n multiplied by m + 1 raise the power of 2. So if we group up the two n exponents together, we will get this 2n, but we will be left with a multiple of n + 1 squared. Now for the Y terms here, in the numeratory to the power of M + 1 will simplify to be Y to the power of N multiplied by Y, and this will be divided by Y to the power of N. And this term is going to further simplify to just be why. So, what we are left with is the limit, as N approaches infinity of the absolute value of the quantity N + 1 divided by N raises to the power of 2n multiplied by N + 1 quantity squared multiplied by Y. And if we were to apply the limit, we will end up with positive infinity. No, What this infinity result tells us is that we are going to be infinitely convergent, and when we have a result of infinity when finding the radius and interval of convergence, then our radius of convergence is going to be centered at 0 with the interval of convergence being the singlesin set 0. And this is going to be the solution to the problem. And with that being said, the overall solution is going to be B. So I hope this video helps you in understanding how to approach this problem, 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₌₁∞ (kx)ᵏ

Key Concepts

Power Series

Radius of Convergence

Interval of Convergence

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