Power Series
In Exercises 47–56, (a) find the series’ ra...

Question

Power Series

In Exercises 47–56, (a) find the series’ radius and interval of convergence. Then identify the values of x for which the series converges (b) absolutely and (c) conditionally.

∑ (from n = 1 to ∞) (x + 4)ⁿ/(n3ⁿ)

Accepted Answer

Hello there. Today we are going to solve the following practice problem together. So first off, let us read the problem and highlight all the key pieces of information that we need to use in order to solve this problem. For the power series, the sum evaluated from n equals 1 to infinity of parentheses of x minus 2 to the power of n divided by 5 n multiplied by n. A Find the radius R and interval of convergence, and the values of X for which the series converges B absolutely and C conditionally. OK, so it appears for this particular prompt we are given a power series, and using this power series we are asked to solve for 3 separate parts. For Part A we need to find the radius R and the interval of convergence, and for Parts B and C we are asked to solve for the values of x for which the series converges absolutely for Part B and where the series converges conditionally for Part C. Fantastic. So with that in mind, we must first let a subscript N, A for short, be equal to parentheses of X minus 2 to the power of n. Divided by 5 to the power of n multiplied by n. And we need to recall a note that since the general term involves an exponential term, as a result, we can recall and apply the ratio test. So based on our prior knowledge, we should be able to recall that the ratio test states the following. So, the ratio test tells us that if we are given a series, so if we're given a series, the sum of A N. And we define l as being equal to the limit as n approaches infinity of the absolute value of a subscript n plus 1 divided by a n. We need to note that if l is less than 1, the series will converge absolutely. And if l is greater than 1 or if l is equal to infinity, that will mean that the series diverges. And if L is equal to 1, that will mean that our test will be inconclusive. So with that in mind, we now need to solve for a subscript n + 1, and when we find a subscript n + 1, we will find that it's equal to parentheses of x minus 2 to the power of n + 1 divided by 5 to the power of n + 1 multiplied by n + 1 in parentheses. Now we need to find the ratio, so we need to take the absolute value of a subscript N + 1 divided by AN. Once again, don't forget your absolute value signs, which will be equal to the absolute value of parentheses of X minus 2 to the power of n + 1. Divided by 5 to the power of n + 1 multiplied by parentheses of n + 1 divided by parentheses of x minus 2 to the power of n divided by 5 to the power of n multiplied by n. Which will be ultimately equal to the absolute value of X minus 2 divided by 5, multiplied by N divided by N + 1. So now we need to find our limit, so L will be equal to the limit as n approaches infinity of the absolute value of x minus 2 divided by 5 multiplied by n divided by n plus 1. And this will be ultimately equal to, once we perform our evaluation, it'll be equal to the absolute value of X minus 2 divided by 5, multiplied by N divided by N + 1. Fantastic. So now we can begin to solve for part A, which we need to find the radius and the interval of convergence. Recalling once again that L is less than 1, so if L is less than 1, the series converges absolutely. Thus, so let's make a note, thus that will mean that the absolute value of X minus 2 divided by 5 is less than 1, which will mean the absolute value of X minus 2 will be less than 5. Which will give us a radius of convergence, R is equal to 5, so let's put a box around that answer because that's one of our final answers we're ultimately trying to solve for. And we have a center at x0 is equal to 2. Which will mean as a result the open interval of convergence will be -5 less than x minus 2, which will be less than 5. Which will mean that -3 is less than X and X is less than 7. So now we need to test our endpoints. X is equal to -3, and X is equal to 7, and we are testing these endpoints for convergence by substituting them back into our given series. So starting at x is equal to -3. That will mean, as a result, the sum evaluated from n equals 1 to infinity of parentheses of 3 minus 2 to the power of n. Divided by 5 to the power of n multiplied by n will be ultimately equal to the sum evaluated from n equals 1 to infinity of -1 to the power of n divided by n. So at this point, we need to recall and apply the alternating series test in order to determine convergence. And as we should recall, the alternating series test, also known as the Levine's test, requires us to write the series as the sum of -1 to the power of n multiplied by bn. With bn being greater than or equal to 0 and that bn is greater than or equal to b subscript n + 1, where bn is non-increasing and that the limit as n approaches infinity of bn is equal to zero and the series will converge if all of these conditions are satisfied, otherwise the series will diverge. So with that in mind for the series, the sum evaluated from n equals 1 to infinity of -1 to the power of n divided by n, that will mean that BN is equal to 1 divided by n. We also need to note that. BN so BN. Is going to be greater than or equal to 0, so that will mean that as a result, bn will be equal to 1 divided by n, which will be greater than or equal to b subscript n + 1, which will be equal to 1 divided by n + 1, where bn is non-increasing and the limit. As n approaches infinity of bn is equal to the limit, as n approaches infinity of 1 divided by n is equal to 0. And since all of these conditions are satisfied, that means the series converges, so I put C in quotation marks. And similarly, at x is equal to 7, that will mean as a result, the sum evaluated from n equals 1 to infinity of parentheses of 7 minus 2 to the power of n divided by 5 to the power of n multiplied by n. Will be ultimately equal to the sum evaluated from n equals 1 to infinity of 1 divided by n. And this is a P series, and we need to note that this is a P series with P is equal to 1, which is less than or equal to 1, thus our series converge. So once again, our series will diverge in this case because of this P series case. So now that we know that the series diverges, we could therefore state without a doubt that our interval of convergence will be square bracket of 3.7, and let's put a box around that answer because that's one of our final answers that we are asked to solve for. Awesome. Now we can solve for Part B. So moving on to solving for absolute convergence, we need to find the absolute series first, so the sum evaluated from n equals 1 to infinity of the absolute value of parentheses of x minus 2 to the power of n. Will be divided by 5 to the power of n multiplied by n and don't forget your absolute value sign, and this will be equal to the sum evaluated from n equals 1 to infinity of the absolute value of x minus 2 divided by 5 to the power of n multiplied by 1 divided by n. And if we let the absolute value of x minus 2 divided by 5 be less than 1, that will mean, that will mean as a result, -3 will be less than x and X will be less than 7. So at both endpoints, the absolute series will be the sum evaluated from N equals 1 to infinity of 1 divided by n, and this will be a divergent P series with P being equal to 1 and 1 is less than or equal to 1. Thus, the interval where the series converges absolutely will be parentheses of -3. 7 in parentheses, and this is one of our final answers so we could put a box around it. So now we can solve for Part C, which is our final answer. And we need to recall a note that sense only endpoint, so the only endpoint where the series converges, but it is not absolutely converging, will be at x is equal to -3. Thus, the series converges conditionally at x is equal to -3. And that's our final answer. Hooray, we did it. So going back to the top to recap what we've learned, we can safely say that our final answers for parts A, B, and C are as follows. So for Part A, we know that R is equal to 5, and we know that our interval of convergence is square bracket of -3.7. And for Part B, we know that it's parentheses of 3.7. And we know that C is X equal to -3, and that's it. Thank you so much for watching, hopefully that helped, and I can't wait to see you in the next video. Bye.

Power SeriesIn Exercises 47–56, (a) find the series’ radius and interval of convergence. Then identify the values of x for which the series converges (b) absolutely and (c) conditionally.∑ (from n = 1 to ∞) (x + 4)ⁿ/(n3ⁿ)

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Power Series
In Exercises 47–56, (a) find the series’ radius and interval of convergence. Then identify the values of x for which the series converges (b) absolutely and (c) conditionally.

∑ (from n = 1 to ∞) (x + 4)ⁿ/(n3ⁿ)