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 (b) absolutely?

∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)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. Determine the interval of absolute convergence for the series. The sum evaluated from n equals 0 to infinity of 3 to the power of n multiplied by parentheses of n + 1 multiplied by parentheses of x + 2 to the power of n. OK. So it appears for this particular problem, we are simply asked to take the series that is provided to us by the prom itself, and we're asked to ultimately determine what the interval of absolute convergence is for this specific series. So, once again, what is the interval of absolute convergence for this specific series? So now that we know that we're ultimately trying to solve for the interval of absolute convergence for this specific series, we need to start by finding the absolute series. So we need to take the sum evaluated from n equals 0 to infinity of the absolute value of parentheses of 3 to the power of n multiplied by parentheses of n + 1. Multiplied by parentheses of x + 2 to the power of n. And don't forget your absolute value sign, which will be equal to the sum evaluated from n equals 0 to infinity of parentheses of 3 multiplied by the absolute value of x + 2. And don't forget your parenthesis to the power of N. Multiplied by parentheses of n + 1. So, our next step now is we need to let the value a subscript NN for short, be equal to parentheses of 3 multiplied by the absolute value of x + 2. To the power of n. Multiplied by parentheses of n + 1. And we need to note that since the general term involves an exponential term, we can recall and apply the ratio test. So, let's make a note of that. So, once again, because we note that the general term involves an exponential term, and because it involves an exponential term, we can recall and use the ratio test as a result. So what is the ratio test? Well, as we should recall, the ratio test states that if we are given a series, so given a series such as the sum of a n, we can define L is 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 means 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 with that in mind, our next step now is we need to find the value of a subscript N + 1. So a subscript N + 1 will be equal to parentheses of 3 multiplied by the absolute value of x + 2 to the power of n + 1 multiplied by parentheses of n + 1 + 1. Which will be equal to parentheses of 3 multiplied by the absolute value of x + 2 to the power of n + 1 multiplied by parentheses of n + 2. Fantastic. Now, at this point, we need to determine or to find the ratio, so we need to recognize that both A N and a subscript N + 1 are positive terms. So we need to take the absolute value of a subscript N + 1 divided by AN, which will be equal to parentheses of. 3 multiplied by the absolute value of x + 2 to the power of n + 1 multiplied by n + 2 in parentheses divided by parentheses of 3 multiplied by the absolute value of x + 2 to the power of n multiplied by parentheses of n + 1. And when we write our final answer in its most simplified form, meaning we can't manipulate our expression or simplify it any further, we will find that our answer will be 3 multiplied by the absolute value of x + 2 multiplied by n + 2 divided by n + 1. So now, at this stage, we need to find the limit. So let us note that L will be equal to the limit as N approaches infinity of 3 multiplied by the absolute value of X + 2 multiplied by N + 2 divided by N + 1. Which will be equal to 3 multiplied by the absolute value of x + 2 multiplied by the limit as n approaches infinity of 1 + 2 divided by N divided by 1 + 1 divided by n. Which will be equal to 3 multiplied by the absolute value of x + 2 multiplied by square brackets of 1 + 0 divided by 1 + 0, which will be equal to 3 multiplied by the absolute value of x + 2. So now, at this stage, we need to find the radius and the interval of convergence, and once again, we need to note that the series will converge absolutely if L is less than 1. So it's gonna converge absolutely if L is less than 1, and that's what C.A represents is converges absolutely. So considering that fact, we need to note thus that 3 multiplied by the absolute value of x + 2 is less than 1, which will mean that the absolute value of x + 2 is less than 1 divided by 3. And this will give us a radius of convergence. R is equal to 1/3, and this will be centered at X0 equal to -2. So I'll put a note that this is centered. At X0 is equal to 2. Fantastic. So, moving right along here, therefore, as a result, the open interval of convergence will be 1/3 is less than x + 2, which is less than 1/3, which will mean 7 divided by 3 will be less than x and X will be less than 5 divided by 3. Now we need to test our endpoints. X is equal to -7 divided by 3, and x is equal to -5 divided by 3, and we need to test these endpoints for absolute convergence by substituting them into our obtained absolute series. And since we need to recognize that both endpoints are obtained from solving, so once again, because once again these endpoints are both obtained from solving the absolute value of X + 2. Is equal to 1/3, that will mean that substituting both the endpoint values into our absolute series will yield the following result, which will give us the sum evaluated from n equals 0 to infinity of parentheses of 3 multiplied by 1/3 to the power of n multiplied by n + 1. So N + 1. And we will need to note that this will be equal to the sum evaluate from n equals 0 to infinity of 1 to the power of n multiplied by parentheses of n + 1, which will be all equal to the sum evaluated from n equals 0 to infinity of parentheses n + 1. Fantastic. So with that in mind, we now need to recall and apply the nth term test for divergence, and we need to note that if the limit as n approaches infinity of c subscript n. Does not exist or is not zero, then that will mean as a result, the sum evaluated from n equals 0 to infinity of c subscript n or cn will diverge as a result. So I put ad in quotation marks to represent divergence. So we need to note that for the series, which is the sum evaluated from N equals 0 to infinity of N + 1, that will mean that CN will be equal to N + 1. Then that will mean as a result that the limit as n approaches infinity of n + 1 in parentheses will be equal to square brackets of infinity plus 1, which will be equal to infinity and infinity cannot be equal to zero. Thus, as a result, the series diverges. Therefore, without a doubt, there is no absolute convergence, so there's no absolute convergence. For both end points, so once again there is no absolute convergence for both end points as a result. Therefore, the interval of absolute convergence will be parentheses of 7 divided by 35 divided by 3, and that's it. That is our final answer. Hooray, we did it. So going back to the talk to recap what we've learned, we can safely say that our final answer, once again, without a doubt, will be parentheses of 7 divided by 35 divided by 3, and that's it. That is our final answer. 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, for what values of x does the series converge (b) absolutely?∑ (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 (b) absolutely?
∑ (from n = 0 to ∞) [ (−2)ⁿ (n + 1) (x − 1)ⁿ ]