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 ∞) [ (x² − 1) / 2 ]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. 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 0 to infinity of parentheses of x 2 minus 2 divided by 7 all to the power of n. OK, So it appears for this particular problem that we're ultimately asked to solve for two separate answers. So, our first answer we are asked to solve for is the interval of convergence, and our second answer we're asked to solve for is the sum as a function of x. And we're asked to solve for this, both of these pieces of information by using our series that is provided to us. So using the series, we will solve for the interval of convergence and the sum as a function of X. So now that we know what we're ultimately trying to solve for, let us state that a a subscriptn will be equal to parentheses of x 2 minus 2 divided by 7, all to the power of n. And we need to be able to recall and recognize that this is a geometric series with the first term 1, and our common ratio, which will denote as lowercase r, and the common ratio r will be equal to x 2 minus 2 divided by 7. And we need to recall a note based on our prior knowledge, a geometric series will converge. If the absolute value of R is less than 1, so once again for a geometric series, if the absolute value of R is less than 1, it will converge, so I put a C in quotation marks. So that means we can go ahead and write that the absolute value of x 2 minus 2 divided by 7. So once again, the absolute value of x 2 minus 2 divided by 7, will be less than 1, which means we can go ahead and write that the absolute value of x 2 minus 2 will be less than 7. So that will mean that 7 is less than x 2 minus 2, which is less than 7. Which means that -5 is less than X2, which is less than 9. So that means that -5 is less than X 2 is always true for all X. Thus, we only need to focus our attention on, so let's make a note, we only need to focus on, or we only need X 2 is less than 9. So therefore, the open interval of convergence will be the absolute value of X is less than 3, which will mean that -3 is less than X and X is less than 3. So now, at this point, we need to test our endpoints, which is X is equal to -3 and X is equal to 3. And we need to test our endpoints for convergence by substituting them into our given series starting with at x is equal to -3. That will mean that our series, the sum evaluated from n is equal to 0 to infinity of parentheses -3 to the power 2 minus 2 divided by 7 all to the power of n. Will be ultimately equal to, when we go through and do our math, we will find that it's ultimately equal to the sum evaluated from. N is equal to 0 to infinity of 1. So once again we just expand our expression and simplify it down to its most simple form and then perform the evaluation. And when we do, we will find that it's simply equal to the sum evaluated from n is equal to 0 to infinity of 1. So I didn't do the math for you, this is just simply, you're just expanding it, simplifying it, and then evaluating it. So it's just a little bit of algebra basic math. So. Similarly, for x is equal to 3, we can write that the sum evaluated from n is equal to 0 to infinity of parentheses of 32 minus 2 divided by 7. All to the power of n will be equal to the sum evaluated from n is equal to 0 to infinity of once again, it's going to be of 1. So that will mean at this point, we need to recall and apply the nth term test for divergence. So we need to recall a note that if the limit as n approaches infinity of CN. Does not exist DNE for short. So if this limit does not exist or is not zero, then that will mean as a result that the sum evaluated from n is equal to 0 to infinity of c n diverges as a result. So I put ad in quotation marks. So for the series evaluated from n is equal to 0 to infinity of 1, that will mean as a result that CN is equal to 1. Then that will mean that the limit as n approaches infinity of cn will be equal to the limit as n approaches infinity of 1 and 1 is equal to 1 and 1 cannot equal 0. Thus that means our series evaluated from n is equal to 0 to infinity of 1 also diverges as a result. So therefore, without a doubt, we can simply write that our interval of convergence will be parentheses of 3.3, and this is one of our final answers. Hooray, we did it. So we solve for one answer, now it's time to solve for our second final answer, which is the sum. So we need to solve for S. So based on our prior knowledge, we need to recall and note that the formula for the sum of a convergent series is s, the sum is equal to a subscript 1 divided by 1 minus r, which means we can go ahead and write that the sums is equal to 1 divided by 1 minus x to the 2 minus 2 divided by 7. And when we rearrange this expression to write in its most simplified form, meaning we can no longer manipulate it using a little bit of algebra, you will find that it's simply equal to 7 divided by 9 minus x to the power of 2, and that's it. That is our final answer. Hooray, we did it. We solve for our final two answers. So going back to the top to recap what we've learned, that means we could safely say once again that our final two answers without a doubt will be as follows. So our interval of convergence will be parentheses of 3.3, which indicates an open interval. And our sums will be equal to 7 divided by 9 minus X to the power of 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 ∞) [ (x² − 1) / 2 ]ⁿ

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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 ∞) [ (x² − 1) / 2 ]ⁿ