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Find the values of $ x $ for which the series con…

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Problem 59 Medium Difficulty

Find the values of $ x $ for which the series converges. Find the sum of the series for those values of $ x. $
$ \displaystyle \sum_{n = 0}^{\infty} \frac {(x - 2)^n}{3^n} $

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Calculus 2 / BC

Calculus: Early Transcendentals

Chapter 11

Infinite Sequences and Series

Section 2

Series

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Video Transcript

Let's find the values of X for which the Siri's converges and for these values of X, will actually go ahead and find some as well. So let's break this into two parts. First, we just want to know the X values for which it converges. So let's call that part one and part two's over here when we find the sum. So let's just rewrite this Siri's This is actually a geometric series may be in disguise the way that a certain right now I'Ll just pull out that three and then put the end on the outside of the parentheses. So we have geometric and we see that are equals X minus two over three. And we know that geometrics on ly converge when the absolute value of our is less than one. So we need the absolute value our which, in our case, absolute value of X minus two over three. You simplify that a little bit. Oh, we need that to be less than one due to this over here. So solve this for XO. First, multiply the three over and then using the definition of the absolute value. This means the X minus two satisfies this and then add the tutu all sides of the inequality. So in this case, all three sides and we have our interval for X. So these are the X values for part one for which the series will converge. And we get this from using the fact that the series is geometric and we know geometric on ly ca merges when the absolute value ours lesson one. All right, so that's enough for part one, not for part two geometric series. We know the sum will equal the first term of the series. This is always the formula. The nice thing about this formula here, the way that his friend is it doesn't depend on what the starting point is. So here we're just going we just go to the first term by plugging in the smallest. And that we see in this case is zero. So you have X minus two to the zero power over one minus R. In this case, we know where arias that's just X minus two over three. So we have let me come down here. We have one of top and then one. Let's go ahead and get that common denominators. Three minus X minus two. But in the denominator, I have another three down there. So let me put that three on the top and then let's go ahead and simplify. We have three and then plus two, which is five and then minus X. So for the answer for part two again, we're only assuming that we're only looking at these exiles betweennegative one and five. So assuming exes in there than the value of the summation that we were supposed to evaluate is equal to three over five minus X, and that's your final answer.

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Calculus: Early Transcendentals

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