Intervals of Convergence
In Exercises 1–36, (a) find the...

Question

Intervals of Convergence

In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?

∑ (from n = 0 to ∞) [ n xⁿ / (4ⁿ (n² + 1)) ]

Accepted Answer

Hello. In this video, we are going to be finding the radius of convergence and the interval of convergence for the given series. The series given to us. Is the series from 0 to infinity of n multiplied by x to the power of n divided by 5 n multiplied by n2 + 2. Now, in order to find the radius and interval of convergence, we can go ahead and apply the ratio test. Now the ratio test asks us to do the following. If we are given a series, we want to identify a subn within the series, then we will take the limit as n approaches infinity of the absolute value of a n + 1 divided by a subn. Now the conditions are as followed. The first condition is that if the limit is less than one, the series converges. If the limit is greater than one, the series diverges. And if the limit is equal to one, this is inconclusive. So, what we want to go ahead and do is set up the infinite limit. Now, what AN + 1 is telling us to do is we want to take the value of N + 1 and plug this in for every N within the given series. So, by doing this, we will have the limit as N approaches infinity of the absolute value. Now, we are going to take N + 1 and plug this into our series. So that is going to be N + 1 multiplied by X of N + 1. And this will be divided by 5 raise to the power of N + 1, multiplied by N + 12 + 2. Now here we can create a complex fraction and divide by the original summation, but we can equivalently multiply by the original summations reciprocal. So we will multiply by 5 raises the power of n, multiply by n2 + 2. Divided by n multiplied by x to the power of n. Now there are a few things to simplify in this limit, so let's go ahead and simplify piece by piece. Let's group up the X terms together. So we have X N + 1 divided by X N. Now, by using our exponential properties, we can rewrite the numerator as x to the power of n multiplied by X, and this will divide to be x to the power of n. And x to the power of n will simplify the numerator and denominator, leaving us with just x. Now let's go ahead and group up the constant terms 5 ends together. So we will have 5 n in the numerator, and 5 of n + 1 in the denominator. Again, by using the exponential properties, we can rewrite this as 5 to the power of n divided by 5 to the power of n multiplied by 5, and this will simplify to be 1/5. Next, we will go ahead and group up the end terms together. So we will have And plus 1 Multiplied by n2 + 2. Divided by n + 12 + 2 multiplied by n. Now here if we were to apply the limit, this is just going to simplify to be 1 because the overall power in the numerator is going to be an exponent of 3 and the overall power in the denominator will be an exponent of 3. So if we were to simplify, we can go ahead and rewrite the limit to be the limit as n approaches infinity. Of 1/5 multiplied by x multiplied by n + 1. Multiplied by n2 + 2 divided by n + 12 + 2 multiplied by n. Now, if we were to move the non-n terms to the front of the limit, we can further rewrite the limit to be X, the absolute value of X divided by 5, multiplied by the limit as N approaches infinity. Of n + 1 multiplied by n2 + 2 divided by n + 12 + 2 multiplied by n. Now, as we stated earlier, the overall exponential values of the numerator and denominator are the same. So if we apply our infinite limit properties, this whole limit is going to simplify to be one. So what does this tell us? This tells us that the absolute value of x divided by 5 is going to be less than 1, and if we isolate the x property or the x variable, we can rewrite this inequality to be the absolute value of x less than 5. What this tells us is that our radius of convergence is going to equal to 5, and that is going to be one of the solutions to the problem. Now we need to find our interval of convergence. Now, in order to find the interval of convergence, we first need to determine the end points of the interval. If we rewrite the given inequality, the absolute value of x less than 5 can be rewritten as -5 less than x less than 5. Now, we need to plug in the constants into the original summation and evaluate each of those summations independently. So let's go ahead and start with when X is equal to -5. So if we plug in -5 into our original summation, we get the summation from 0 to infinity. Of n multiplied by -5 raised to the power of n divided by 5 to the power of n multiplied by n2 + 2. Now if we were to further simplify. This can be rewritten. As n Multiplied by -1 to the power of n, multiplied by 5 n, all divided by 5 n multiplied by n2 + 2. And this is going to simplify our summation to be the summation from 0 to infinity. Of -1 to the n multiplied by n divided by n2 + 2. Now what we have here is a convergent series, and it converges by the alternating series test. But since this endpoint converges, we are going to be including this in our interval of convergence. And now we need to repeat this process before the 2nd endpoint. So we're going to test when X is equal to 5. Now when X is equal to 5, we get the summation from 0 to infinity of N multiplied by 5 raised the power of N divided by 5 to the power of N multiplied by N 2 + 2. This is going to simplify the summation to be from 0 to infinity of N divided by N2 + 2. And here we have a series that diverges by the limit comparison test. So since this is divergent, we are not going to include the value of 5 in our interval of convergence. Now that we know that -5 endpoint is convergent and 5 endpoint is convergent or or is divergent, apologies, our interval of convergence is going to be from -5 included to 5 not included, and that is going to be the second solution to this problem. And now that we've identified the radius of convergence, the interval of convergence, the overall solution to this problem is going to be option A. So I hope this video helps you in understanding on how to approach this problem, and we will go ahead and see you all in the next video.

Intervals of ConvergenceIn Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?∑ (from n = 0 to ∞) [ n xⁿ / (4ⁿ (n² + 1)) ]

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Intervals of Convergence
In Exercises 1–36, (a) find the series’ radius and interval of convergence. For what values of x does the series converge (b) absolutely, (c) conditionally?
∑ (from n = 0 to ∞) [ n xⁿ / (4ⁿ (n² + 1)) ]