In Exercises 43–50, use Theorem 20 to find the series’ interva...

Question

In Exercises 43–50, use Theorem 20 to find the series’ interval of convergence and, within this interval, the sum of the series as a function of x.

∑ (from n = 0 to ∞) [(x + 1)²ⁿ] / 9ⁿ

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. Find the interval of convergence, and within that interval the sum as a function of x of the series, the sum evaluated from n is equal to 1 to infinity of parentheses of x minus 2 to the power of 2 n divided by 16 to the power of n. OK, so it appears for this particular prom that we're asked to ultimately solve for two separate answers. Our first answer that we're asked to solve for is the interval of convergence. And our second answer we're ultimately asked to solve for is the answer for the sum. So we're solving for the sum denoted ass. So what does s equal? And we're asked to determine what the interval of convergence and the sum is for this specific series that is provided to us. So now that we know what we're ultimately trying to solve for, let us be able to recall and recognize that the series that is given to us. Can be written as the sum evaluated from n is equal to 0 to infinity of parentheses of x minus 2 to the power of 2 n divided by 16 to the power of n. is equal to the sum evaluated from n is equal to 0 to infinity of parentheses of x minus 2 to the power of 2. Divided by 16, all to the power of n. So once again we can rewrite our series as such. So, at this stage, it is very important that we're able to recognize that this is a geometric series with the first term one. And a common ratio which will denote the common ratio as lowercase r and r, the common ratio is equal to parentheses of x minus 2 to 2 divided by 16. Fantastic. So moving right along here, we also need to recall and note that a geometric series converges if. The absolute value of R is less than 1. So if the absolute value of R is less than 1, that will mean it will converge. So I put a C in quotation marks so you can remember. So once again, if the absolute value of R is less than 1, that means it will converge for a geometric series. So with that said, we can go ahead and write that the absolute value of the parentheses of x minus 2 to the power of 2 divided by 16. is less than 1, which will mean you can write that parentheses of x minus 2 to 2 divided by 16 is less than 1, which will mean that parentheses of x minus 2 to 2 is less than 16, which will mean that the absolute value of x minus 2. So once again, the absolute value of X minus 2 will be less than 4. And this will give us the radius of convergence, and the radius of convergence is denoted by capital R, and the radius of convergence is equal to 4, and we need to note that the center is at x0x subscript 0 is equal to 2. Fantastic. So that means as a result, the open interval of convergence is -4 less than x minus 2, which is less than 4, which will mean we can write 2 is less than x and X is less than 6. So now at this point, we need to test our endpoints. X is equal to -2 and X is equal to 6, and we need to test these endpoints for convergence by substituting them into our. G series. So starting with at x is equal to -2, the series is the sum evaluated from n is equal to 0 to infinity of parentheses of parentheses of 2 minus 2 to the power of 2 divided by 16 to the power of n. Which will be equal to the sum evaluated from n is equal to 0 to infinity of parentheses of. 4 in parentheses to the power of 2 divided by 16. All to the power of n. And when we continue this pattern and we have our answer written in its most simple form, which I'm going to skip a few steps, we will find that this is simply equal to the sum evaluated from N is equal to 0 to infinity of 1. So all we're doing is just performing our math until we write our final answer in its most simplified form and perform the evaluation. And so when we do that, we will find that it's simply equal to the sum evaluated from n is equal to 0 to infinity of 1. Similarly, for the at x is equal to 6 condition, that will mean we could write that the sum evaluated from n is equal to 0 to infinity of parentheses of parentheses of 6 minus 2 to the power of 2 divided by 16 all to the power of n. This will be ultimately equal to the sum evaluated from n is equal to 0 to infinity of simply 1 again. Fantastic. So now at this point, we need to recall and apply the nth term test for divergence. So as we should recall. If the, so once again, if the limit as n approaches infinity of Cn does not exist, DNE for short, or is not zero, then we can write, so once again, then. The sum evaluated from n is equal to 0 to infinity of CN will diverge as a result. So I put D in quotation marks. So, for our series, so for the series, the sum evaluated from N is equal to 0 to infinity of 1, that will mean that CN is equal to 1. Then as a result that will mean that the limit as n approaches infinity of cn will be equal to the limit as n approaches infinity of 1, which is equal to 1 which cannot equal 0. Thus, as a result, our series, the sum evaluated from n is equal to 0 to infinity of 1 will. Undoubtedly diverge, so it diverges. So I put I put ad in quotation marks. So therefore, without a doubt, our interval of convergence will be parentheses -2.6, and that's it. That is one of our final answers. Hooray, we did it. Now we need to solve for the sum, which is our second and final answer. So switching gears a little bit, we need to note and recall based on our prior knowledge the sum of a convergent geometric series formula is written as the sums is equal to a subscript 1 divided by 1 minus r. Which in this case for the case of our problems will be equal to 1 divided by 1 minus parentheses of x minus 22 divided by 16. Which will mean when we rearrange our expression to write our final answer in its most simplified form, meaning we can't simplify it any further or manipulate it more, we will find that it's equal to 16 divided by 12 + 4 x minus x to the power of 2, and that's it. That is our final answer for the sum. Hooray, we did it. So going back to the top to recap what we've learned, we can safely say that our final two answers for this particular problem are as follows. So let us note that the interval of convergence once again is parentheses of -2.6. And the sums is equal to 16 divided by 12 plus 4 x minus x 2. And that's it. Those are our final two answers. Thank you so much for watching. Hopefully that helped, and I can't wait to see you in the next video. Bye.

In Exercises 43–50, use Theorem 20 to find the series’ interval of convergence and, within this interval, the sum of the series as a function of x.∑ (from n = 0 to ∞) [(x + 1)²ⁿ] / 9ⁿ

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In Exercises 43–50, use Theorem 20 to find the series’ interval of convergence and, within this interval, the sum of the series as a function of x.
∑ (from n = 0 to ∞) [(x + 1)²ⁿ] / 9ⁿ