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∞ ((−1)ᵏ⁺¹(x−1)ᵏ)/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. The power series given to us is the infinite series starting at 1 of X + 2 quantity rates to the end power, divided by N multiplied by 3 rates to the nth power. Now, in order to find the radius of convergence, we can go ahead and apply the ratio test. The ratio test asks us to take the limit, as N approaches infinity of the ratio AN + 1 divided by AN. So we're going to need to plug in the quantity M +1 into the original argument of the series, then we want to divide that quantity by the original argument. So, by applying the ratio test, we will have the limit as N approaches infinity of the absolute value of the quantity X + 2, raise to the power of N + 1. Divided by the quantity M + 1, multiplied by 3, raise the power of N + 1. Now, here, we will need to divide this by the original argument that is given to us in the series, but that's going to create a confusing looking complex fraction. So, equivalently, what we can do is we can multiply by the reciprocal of the original argument. So that is going to be N multiplied by 3, rates the power of N divided by the quantity, X plus 2, rates the power of N. And so now what we want to go ahead and do is we want to simplify. Let's go ahead and start by simplifying the X + 2 terms. Now in the limit in the numerator, we have X + 2, raise the power of M + 1, and in the denominator, we have X + 2 raise the power of N. We can express this numerator as a product of exponents and rewrite it as X + 2, raise to the power of N multiplied by X + 2, and this will be divided by the quantity X + 2 raises to the power of N. We can eliminate the X + 2 rate of the power of N in both the numerator and denominator, and that will leave us with the quantity X + 2. So this is going to be the X + 2 term simplified. Now, what about the N terms in this case? Well, in the numerator, we have N, and in the denominator we have N + 1. There's nothing further that we can do to simplify this, so this is in the simplest form. And finally, what about the three N terms? Well, in the numerator, we have 3 to the power of N, and in the denominator, we have 3 to the power of N + 1. We can apply the same trick that we did in the X + 2 terms and rewrite this denominator as 3 to the power of N multiplied by 3, and this is going to simplify to leave us with 1/3. So by putting this back into the limit and multiplying some terms together, we can rewrite the limit to be the limit as N approaches infinity of X + 2 divided by 3, multiplied by N divided by N + 1. Then if we were to move X + 2 divided by 3 as a constant in front of the limit, we can further rewrite the limit to be the absolute value of X + 2 divided by 3, multiplied by the limit as N approaches infinity of N divided by N + 1. And this limit is going to simplify to be zero, so the output of the ratio test is going to be the quantity, absolute value of X + 2 divided by 3. Now, in order for the ratio test to be convergent, we want our limit to be less than one. So we are going to set the quantity X + 2 divided by 3 less than 1, and what we want to do is we want to go ahead and isolate this X + 2 term. In order to do this, we will multiply both sides of this inequality by 3, and that is going to leave us with X + 2 less than 3. And so here, since we have the X term isolated in the absolute value, we want to take a look at the constant term on the right hand side of the inequality. This constant term tells us that the radius of convergence for the series is going to be 3. And so now that we have the radius of convergence, we want to go ahead and identify the interval of convergence. Now, if we were to expand this inequality, we can rewrite the inequality to be X + 2 is trapped between the values of -3 and 3, and by isolating the value of X in the inequality, we will have that X is between 5 and 1. So that tells us that the end points of our interval of convergence is going to be -5 and 1, but now we need to check to see which of these endpoints is going to be convergent. So, in order to do this, we are going to take both values of the end points, plug them back into the original power series, and determine the convergence of each of them. So starting with X is equal to -5, if we were to plug this back into the original power series, we have the power series starting at 1 of -5 + 3 raise to the N power, which is going to be -3, raise to the power of N divided by N, multiplied by 3 N. And this is going to further simplify to be the power series starting at one. Of -1 to the nth power. Divided by n. Now, if we were to apply the alternating series test, The conclusion that we will get is that this is going to be a convergent alternating series. And so that means that at X is equal to -5, we are going to have a convergent endpoint. But now we need to check when X is equal to 1. If we were to plug in positive 1 back into the original power series, we will have 1 + 2, which is going to be 3 power of N divided by N, multiplied by 3 to the power of N. This is going to further simplify to give us an infinite power series starting at 1 of 1 divided by N. This is going to be a divergent harmonic series. And so that means that for our interval of convergence, we will not be including the endpoint of positive 1. So just to recap, we went ahead and plugged in both endpoints of our interval of convergence into the original power series. We determined that at -5, we have a convergent endpoint, and at 1, we have a divergent endpoint. And so what that means is that the interval of convergence is going to be from -5 to 1, not including 1. And just to recap from earlier, by applying the ratio test, we were able to find that the radius of convergence is going to be 3, and these are going to be the solutions to the problem, with the overall solution being A. 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∞ ((−1)ᵏ⁺¹(x−1)ᵏ)/k

Key Concepts

Power Series

Radius of Convergence

Interval of Convergence

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