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 = 1 to ∞) [ (x − 1)ⁿ / √n ]

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 series, the sum evaluated from n equals 0 to infinity of parentheses of x minus 2 to the power of n divided by the square root of n + 1. A Find the radius R and the interval of convergence, and state where it converges B absolutely and C conditionally. OK. So it appears for this particular prompt. We are given a series, and using this series we're asked to solve for 3 separate parts. For Part A, we are asked to find the radius R and the interval of convergence, and for Parts B and C, we are asked to state where it converges absolutely for Part B and where it converges conditionally for Part C. OK. So now we know what we're ultimately trying to solve for. Our first step is we need to let An be equal to parentheses of x minus 2 to the power of n divided by the square root of n + 1. And since the general term involves an exponential term, that means as a result, we can recall and apply the ratio test. So with that in mind, we need to recall and note the following. So to take a moment here to pause, so we need to note that we are given a series. And we need to note. That with this given series, the sum of a subscript n, then you need to define l as being equal to the limit as n approaches infinity of the absolute value of a subscript n + 1 divided by a n. And if L is less than 1, that will mean that the series converges absolutely. And if L is greater than 1, or L is equal to infinity, that will mean the series diverges, and if L is equal to 1, that will mean that the test will be inconclusive. So, keeping the ratio test in mind, and let's make a note up top that we have, that we, this is our rules for the, I would say more like conditions instead of rules. These are the conditions for the ratio test. We will now need to find a subscript N + 1. So a subscript N + 1 will be equal to parentheses of X minus 2 to the power of N + 1. Divided by the square root of n + 1. Plus one And this will be equal to parentheses of x minus 2 to the power of n + 1 divided by the square root of n + 2. Fantastic. We now need to find the ratio, so we need to take the absolute value of a subscript N + 1 divided by AN. And this will be equal to the absolute value of parentheses of x minus 2 to the power of n + 1 divided by the square root of n + 2 divided by. Parentheses of x minus 2 to the power of n divided by the square root of n + 1. And when we solve for this expression, writing our final answer in its most simplified form, which I'm going to skip a few steps because you need to be, you should know how to manipulate this expression algebraically, we will find that it's equal to the absolute value of x minus 2 multiplied by the square root of n + 1 divided by n + 2. And we need to note that the square root of N + 1 divided by the square root of N + 2 will be equal to the square root of N + 1, so it's gonna be equal to the square root of N + 1 divided by N + 2. And this is the case since both the square root of N + 1 and the square root of N + 2 are valid. 4 N is greater than or equal to 0. So, once again, this is valid, 4 N is greater than or equal to 0. So we now need to find our limit, so L is equal to the limit as n approaches infinity of the absolute value of x minus 2 multiplied by the square root of n + 1 divided by n + 2. Which will be equal to the absolute value of x minus 2 multiplied by the limit as n approaches infinity of the square root of 1 + 1 divided by n divided by 1 + 2 divided by n. And we need to note that as N approaches infinity, that will mean that as a result, 1 divided by N and 2 divided by N will also approach 0. So therefore we can go ahead and write that l is equal to the absolute value of x minus 2 multiplied by square brackets of 1 divided by 1, which will be equal to the absolute value of x minus 2. So now we can begin to solve for Part A. So we're solving for Part A first, which once again we need to find the radius and the interval of convergence. So we need to note that our series will converge absolutely, which I'll call CA for short, and it will converge absolutely if L is less than 1. Thus, we need to write that as a result, the absolute value of X minus 2 must be less than 1. And this will give us a radius of convergence. R is equal to 1, and we will have a center. At X 0 is equal to 2. So let's put a box around R is equal to 1, because that is one of our final answers we are asked to solve for. And based on this information, we could write that the open interval of convergence will be -1 is less than x minus 2, which is less than 1. Which will mean we can write that 1 is less than X and X is less than 3. So now we need to test the end points. X is equal to 1, and x is equal to 3 for convergence by substituting them into our given series. So for the at x equals 1 condition, our series will state the sum, evaluate from n equals 0 to infinity of parentheses of 1 minus 2 in parentheses to the power of n. Divided by the square root of n + 1 will be equal to the sum evaluated from n equals 0 to infinity of parentheses of 1 to the power of n divided by the square root of n + 1. So now at this stage, 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 writing 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 BN being greater than or equal to B subscript N + 1. Where we need to note that BN is a non-increasing value, and we need to note that the limit as N approaches infinity of BN will be equal to 0. And the series converges if all of these conditions are satisfied, otherwise, the series will diverge. So for our series, once again, the sum evaluated from n equals 0 to infinity of -1 to the power of n divided by the square root of n + 1. We need to note that BN is equal to 1 divided by the square root of n + 1. Multiplied by BN, which will be greater than or equal to 0. So that will mean that BN is equal to 1 divided by the square root of, so once again, BN will be equal to 1 divided by the square root of N + 1, which will be greater than or equal to B subscript N + 1, which will be equal to 1 divided by the square root of N + 2. So we need to note that once again 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 the square root of n + 1, which will be equal to square brackets of 1 divided by the square root of. Infinity plus 1, which will be equal to 0. And since all the conditions are satisfied, that will mean that our series converges. Hurray. So, we need to note that for at, so now we're considering the at X equals 3 condition, that will mean, as a result, the series, the sum evaluated from N equals 0 to infinity of parentheses of 3 minus 2 to the power of n divided by the square root of n + 1. Will be equal to the sum. Evaluated from N equals 0 to infinity of 1 to the power of N divided by, so it's gonna be 1 of N divided by the square root of N + 1. And this will be equal to the sum evaluated from N equals 0 to infinity of 1 divided by parentheses of n + 1 to the power of 1/2. Fantastic. So with that said, we need to note that our series, the sum evaluated from n equals 0 to infinity of 1 divided by parentheses of n + 1 to the power of 1/22 behaves like the sum evaluated from n equals 0 to infinity of 1 divided by n to the power of 1/2. And this is gonna behave like a divergent, which I'll put a D in quotation marks. So this is a divergent P series, which I'll call PSense. Let's make a note, since 1/2 is less than or equal to 1. Thus, the series, the sum evaluated from n equals 0 to infinity of 1 divided by parentheses of n + 1 to the power of 1/2. Diverges. So I'll put a d in quotation marks. So therefore, as a result, our interval of convergence will be square bracket of 1.3 parentheses. And let's put a box around this answer because that is one of our final answers that we are asked to solve for. Now we can solve for Part B, and Part B is solving for absolute convergence, so we need to consider the sum evaluated from n equals 0 to infinity of the absolute value of parentheses of x minus 2 to the power of n. Divided by the square root of n + 1, which will be equal to the sum evaluated from n equals 0 to infinity of the absolute value of x minus 2 to the power of n divided by the square root of n + 1. And we need to find the absolute series, which is the absolute value of X minus 2 is less than 1, which will mean that 1 is less than X and X is less than 3. So that means that the absolute series at both end points will be the sum evaluated from n equals 0 to infinity of 1 to the power of n divided by the square root of n + 1 will be equal to the sum evaluated from n equals 0 to infinity of. One divided by. So it's gonna be one. Divided by parentheses n + 1 to the power of 1/22 diverges. So, moving right along here, that will mean that thus, as a result, our series converges absolutely on parentheses 1.3. And let's put a box around that answer because that's one of our final answers we are ultimately asked to solve for. Now we could begin to solve for Part C, and we need to consider conditional convergence, and we need to note that only the only endpoint where the series converges, but it is not absolute convergence, so it's not absolutely converging, is at X equals 1. Thus, the series converges conditionally at. X is equal to 1, and that is our last answer that we were asked to solve for. Hooray, we did it. So going back to the top to recap what we've learned, we can safely say, without a doubt, based on our prior knowledge and what we've learned together, that our final answers for Parts A, B, and C will be as follows. So, our answer for part A is R is equal to 1, and our interval of convergence is square bracket of 1.3, B is parentheses 1.3, and C is X is equal to 1. 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.

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 = 1 to ∞) [ (x − 1)ⁿ / √n ]

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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 = 1 to ∞) [ (x − 1)ⁿ / √n ]