WEBVTT
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okay for this problem will use the ratio test toe
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figure out when we get convergence here for the ratio
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test we limit is n goes to infinity of a
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n sorry and plus one over and with a in
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terms is this whole thing So including the X values
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here. So that's limit as n goes to infinity
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of absolute value of X minus two to the n
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plus one divided by in plus one squared plus one
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. So that's our and plus one. And we're
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dividing that by a m so multiplying by the reciprocal
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. So we're multiplying by n squared plus one over
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X minus two to the end. Okay, so
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X minus two to the M plus one divided by
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X minus two to the end is just going to
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be X minus two and then we have in squared
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plus one divided by in plus one squared plus one
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. So these up top and down below we have
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degree to polynomial is with leading coefficient one. So
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as n goes to infinity, this is just going
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to go toe one. So this is just going
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to turn out to be absolute value of X minus
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two, and we want for that to be less
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than one. Okay, so absolutely of X minus
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two is going to be less than one when X
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minus two is trapped between minus one and one.
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So if we have that minus one is less than
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X minus two. We can add to to both
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sides and see that that corresponds to one is less
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than X And if we have X minus two is
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less than one we can add to to both sides
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to see that that corresponds to X is less than
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three. Okay, so the interval were working with
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is from one to three. We're not sure whether
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or not we include one and whether or not we
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include three at this point. But the length of
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the interval is three minus one, which is too
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to the radius of convergence is half of that to
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the radius of convergence. R is half of the
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interval of convergence half of the length. The radius
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of convergence is just one, and we need to
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figure out whether or not we include X equals one
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and whether or not we include X equals three for
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the interval of convergence. So in X is equal
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to one. If we plug in one here we
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have minus one to the end, divided by n
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squared plus one. And this is going to converge
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by the alternating signed test. That's good. And
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the other thing we're wondering about is X equals three
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. So if you plug in X equals three here
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, then we get three minus two to the end
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. So that's one to the end. One to
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the end is just one. So we would get
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some just won over and squared, plus one,
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one over in squared plus one is goingto converge because
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one over in squared converges the exploding here is something
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that's bigger than two. So we will get convergence
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here and here. The denominators even bigger, right
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? If we make the denominator bigger, we're gonna
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get something smaller. So since this converges, we
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have to have that this converges. So three works
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. One works. So we include both three and
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one in our interval of convergence