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.

∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]

Accepted Answer

Find the radius R at interval convergence for the series from 0 to infinity of -3 to the n, n + 1, X + 2 rays to the n. Now I've written down the tests we'll use for our radius and interval of convergence. We'll make use of the ratio test first because this involves an exponential term. Now the ratio test says given a series AN define L equals a limit as N approaches infinity of the absolute value of AN + 1 divided by IN. If L is less than 1, it converges absolutely. If L is greater than 1 or L is infinity, series diverges. And if L equals 1, the test is inconclusive. So, let's first make our limit L. We have l equals the limit. As in goes to infinity. Of AN plus 1 divided by AN. We can write this then as a limit as in approaches infinity. Of -3 to the n + 1 multiplied. By n + 2 And X + 2 raised to the N + 1. This is divided. By -3 to the end. N + 1 X + 2 raised to the end. In the absolute value. Now from here we just have some algebra to simplify. Now we notice we have x + 2 to the n + 1 divided by x + 2 to the n. In this case, the exponents will go away, just leaving us. With X + 2 to the first, M + 2 divided by M + 1 stays. And for the same reason with -3. -3 to the end goes away, just leaving us -3 to the 1st. We have the limit as in approaches infinity. Of -3. Multiplied By x + 2 in the absolute value, all multiplied by n + 2 divided by n + 1. And then because we have the absolute value. That -3 just turns into a positive 3. Now we can just find our limit. We have the limit as N goes to infinity, but we do notice we have n + 2 divided by n + 1. We divide both terms by n. We end up getting N divided by N + 2 divided by N, all divided by N divided by N plus 1 divided by N. Which gives us 1 + 2 divided by n divided by 1 + 1 divided by n. This limit just ends up going to one. Meaning we have 3 multiplied by the absolute value of x plus 2 as our value for L. Now this converges if L is less than 1, so we can say 3 absolute value x + 2 is less than 1. Or the opposite value of X + 2 is less than 1/3. That then means are. Is 1/3. A sinner is also negative too. We can then find the open interval of convergence. We have 1/3 is less than x + 2, which is less than 1/3, and when we subtract 2 from both sides, we get -7/3 is less than x, which is less than -5/3. So this is our interval currently, but we need to check. The end point to determine the actual interval. So we'll test the endpoints by substituting into the series, which is where our interim tests and our alternating series tests come into play. Let's say we substitute N. X equals. -7/3 1st. We end up getting the sum Of -3 to the end. N + 1 Multiplied by -7/3 + 2 raised to the n. Now we can simplify this. This is from 0 to infinity, by the way. But we can simplify this. Because we have -3 to the end. And this 7 or 7/3 plus 2, rather. Simplifies to -1/3, so we have -3 to the end multiplied by -1/3 to the end. This means we end up getting. 1 to the n multiplied by n + 1. In our series. And because we have 1 to the end, that value is always 1, we have the series N + 1. Now, by the interim test, if the limit of this series does not exist or is non-zero. This series diverges. Well, this limit is going to be non-zero. It's actually going to be infinity. That means for this endpoint, the series diverges. Now we can check X equals -5/3. We get the series from 0 to infinity. Of -3 to the n n + 1. 5/3 + 2 raised to the end. Now, let's try to simplify this. -5/3 + 2 simplifies to 1/3. So similar to the other one, we end up getting -1 race to the N multiplied by N + 1. Now, to determine this series, we need to use the alternating series test. This tells us that if we have the form -1 to the N or -1 N + 1, with BN right after, And BN is greater than 0, and BN is non-increasing, and the limit is equals 0, this series converges. Otherwise, it diverges. Now we have -1 to the n n + 1. This is greater than 0. But, based on what we have here, this series is actually increasing because of the N + 1. So because this is increasing by alternating series, this also diverges. That means our true interval. Is indeed -7/3 to 5/3. With a radius of 1/3. OK, I hope to help you solve the problem. Thank you for watching. Goodbye.

Intervals of ConvergenceIn Exercises 1–36, (a) find the series’ radius and interval of convergence.∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]

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Intervals of Convergence
In Exercises 1–36, (a) find the series’ radius and interval of convergence.
∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]