WEBVTT
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to use the ratio test to see if this converges
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. So we look at the limit n goes to
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infinity and plus one over and we're an is given
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by this term over here. So first we have
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the numerator and plus one so that'LL be co signed
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And then we have an plus one here and we're
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dividing that by the denominator here and and now we'LL
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go ahead and flip this blue fraction and multiply it
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to the green. And when we do that,
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let's see here. So we'll have co signed and
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plus one over three over another co sign and then
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in factorial and then and plus one factorial. So
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let's simplify be separately. So here, notice that
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and plus one with the factorial that can be written
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as and so would go ahead and cancel those factorial
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is there And also if you look at the's terms
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right here, so co sign pi over three.
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So, for example, if an is one just
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right out of a few terms of the sequence and
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you'LL see, it's it's periodic. We have a
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one af negative one half negative one negative one has
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one half and one and that would keep repeating.
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So that means that when if we divide consecutive terms
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here, well, if you divide this divided by
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that, you get one an absolute value. This
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divided by that would be to an absolute value.
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And let's see, here we have. And that
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looks like a largest fraction that we would get.
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It will be too. So this is less than
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or equal to the limit and goes to infinity two
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. And then we still have our n plus one
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from this term over here so again, with the
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absolute value is always going to be one or two
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. And so we'LL just use upper bound and replace
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this with the two that goes zero on the limit
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, which is less than one. So we conclude
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by this that the Siri's convergence by the ratio test