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