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 ∞) xⁿ/nⁿ

Accepted Answer

For the series, from 1 to infinity of X to the N divided by 2 NN. Find the radius r an interval of convergence and state where it converges absolutely and conditionally. Now, to solve this, we'll make use of the ratio test and the alternating series tests. Let's first use the ratio test. This tells us that given the series A N, define L as the limit as N approaches infinity of the absolute value of A and + 1 divided by A N. If this limit L is less than 1, it converges absolutely. If the limit is greater than one or infinity, it diverges. If the L equals 1, it's inconclusive, and you have to use a different test. Now let's find our ratio. We have AN is X N divided by 2 nn. We can also get A N + 1. Which just replaces n with n + 1. You have X to the n + 1 divided by 2 n + 1. Multiplied by n + 1. Now we can put this in our ratio. Absolute value AN plus 1 divided by AN. This equals x n + 1 divided by 2 n + 1. N + 1 Divided by X to the n divided by 2 nn and the absolute value. Now we can rewrite this. Since we have a fraction of a fraction, we can just use our reciprocals and rewrite this as the following. 2 To the n x multiplied by x to the n. Divided by 2. 2 N X N. This is in the absolute value multiplied. By the absolute value Or rather, just in divided by. And plus 1. This simplifies to x divided by 2. Multiplied by n divided by n + 1. We can move the absolute value bars to just be around the X. Now we just need our limit. L Is the limit as in approaches infinity. Of X divided by 2 with the absolute value multiplied by N divided by n + 1. This gives us absolute value X divided by 2. Of 1 divided by 1 + 1 divided by n with the limit. Which just gives us absolute value x divided by 2 multiplied. By one. Absolute value X divided by 2 is L. Now we need to find the radius and interval of convergence. This converges if L is less than 1. So we can then say L less than 1 turns into absolute value X divided by 2 less than 1. This gives us absolute value X is less than 2, whenever we solve for X. That then makes R equals 2. This then means our open interval convergence is from -2 to 2. To find our true interval of convergence, we have to test the end points by substituting into the series. So We will take x equals -2 and X equals 2. And substitute them into the series. We have X equals -2, we get the series. Of -2 N divided by 2 nn. Now, this simplifies to just give us -1 to the end divided by n. For this one, we have to use the alternating series test to determine its convergence. In this case, BN is 1 divided by N. We want to see if, by the alternating series tests, we have of the form -1 to the NBN. BN Is eventually not increasing. And the limit of BN equals 0 as it approaches infinity. Now, based on what we have here, The limit Of BN does equal 0. And BN is non-increasing. So, this series converges. For x equals -2. Now, for X equals 2. We end up getting the series. From 0 to infinity. Of 1 to the n. Divided by it, which is just 1 divided by n. By P series, this is a harmonic series, which this Always diverges. So that means our true interval of convergence. Is given By -2 with a square bracket. And positive to with a rounded bracket or parenthesis. Now we just need absolute convergence. We can do this by finding the absolute series. But the absolute series is just the harmonic series. So this converges absolutely. On the interval from -2 to 2 parentheses. Now conditional convergence. Because the only endpoint where it converges is at -2 from before. We say this converges conditionally. At X equals -2. We now have the answer to our problem. Our radius is 2. Our interval is from -2 to 2 with a bracket on the -2. The absolute is a -2-2 with parentheses. And it is conditionally convergent at X equals -2. This is finally the answer to our problem. OK, I hope to help you solve it. Thank you for watching. Goodbye.

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 ∞) xⁿ/nⁿ

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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 ∞) xⁿ/nⁿ