Intervals of Convergence
In Exercises 1–36, for what val...

Question

Intervals of Convergence

In Exercises 1–36, for what values of x does the series converge (c) conditionally?

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

Accepted Answer

Determine the values of X, if any, where the series from zero to infinity of -1/2 ton, X + 2 n n + 2 converges conditionally. We have 4 possible answers being X equals -4, X equals 0, X equals 0. X equals -2, or there are no values for which the series converges conditionally. Now, I've put 3 tests on the right side that we're going to need to solve this. Our first test will be the ratio test, but we'll first find the absolute series, so we'll take the absolute value. The absolute series then. It's just the absolute value. Which results in giving us the sum. As it equals 0 to infinity. Of 1 half absolute value X + 2. Raised to the n multiplied by n + 2. So we will Use the ratio test here, which tells us that when given a series AN define L as the limit as in approaches affinity of AN plus 1 divided by AN in the absolute value. If L is less than 1, the series converges absolutely. If L is greater than 1 or L equals infinity, the series diverges. If L is equal to 1, the test is inconclusive. So for our series, we'll take L equals the limit. In a process infinity of a n + 1 divided by a N. We can then rewrite our ratio, the limit and the approaches infinity of the absolute value. We end up getting 1/2 x + 2 raised to the n + 1. Multiplied By N + 3 Divided by 1/2 x + 2 to the n multiplied by n + 2. Now we can just do some algebra here to simplify. This ends up getting 1/2 multiplied by the absolute value x + 2. Multiplied by n + 3 divided by n + 2. With our limit as in approaches infinity. Now let's solve for L. Now we notice here we have n + 3 divided by n + 2. We can rewrite that. By dividing every term by n, giving us n divided by n plus 3 divided by n, all divided by n divided by n + 2 divided by n. This gives us 1 + 3 divided by N divided by 1 + 2 divided by N, which simplifies to give us just One, when taking the limit as it approaches infinity. We now end up getting L. Equals 1/2 absolute value x + 2. Now this converges absolutely when it's less than one. So we will then say. We have 1/2 absolute value, X + 2 is less than 1. Or X + 2 is less than 2. That means then our radius of convergence are. is equals to 2. With the sinner At x equals -2. Now we need to find our interval. The open interval of convergence can be given as -2 is less than X + 2, which is less than 2, which turns into -4 X 0. From here we need to check our end points. We can substitute both values of -4 and 0 and 2. are serious. Now, we first have We'll say X. Equals -4. We ended up getting 1/22 raised to the N multiplied by n + 2, -4 + 2 raised to the end. And this is from N equals 0 to infinity. And we notice -4 + 2 gives us -2. We have -2 to the end there. -1/2 multiplied by -2 gives us 1 to the end. Meaning we just have the sum n equals 0 to infinity. Of n + 2. Now, the nth term divergence test tells us if the limit of a series as it goes to infinity does not exist or is not zero, then the series diverges. Since we have infinity, this series goes to infinity. So, X equals -4 divergence. Let's check x equals 0. We'll get a similar case where you have the sum. Of negative 1/2 raised to the n n + 2. 0 + 2 raised to the end. Which gives us the form of an alternating series when we simplify. Of -1 to the n multiplied by n + 2. Now, the alternating series test tells us that our series BN should be greater than equal to 0. It should be non-increasing, and the limit should equal zero. It converges if all conditions are satisfied, otherwise it diverges. Now, if you look at this, This is actually increasing because of the n + 2. So I'll note that down. Because N + 2 is increasing, One of our conditions fails, which means x equals 0, also diverges. We can then say that there are no values. Or conditional convergence. Because there are no values, if we look at our possible answers, we determine that this answer is answer D. There are no values for which the series converges conditionally. OK, I hope to help me solve the problem. Thank you for watching. Goodbye.

Intervals of ConvergenceIn Exercises 1–36, for what values of x does the series converge (c) conditionally?∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]

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Intervals of Convergence
In Exercises 1–36, for what values of x does the series converge (c) conditionally?
∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]