WEBVTT
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the tea's a graph of the sequence to decide whether
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it converges or diverges, and then from the Graff
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. If it's conversion, will guess the value and
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then we'LL prove our guests. So this one.
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Let's just go ahead and write out a few terms
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. A one is just a half so we could
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put that on the graph. Now a two that's
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just one times three and then here will have four
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squared. So three over sixteen. So notice the
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formula and this up here in the numerator. That's
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the product of the first and our insiders. Positive
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injuries there a three. One times, three times
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five first three odds. Then this time would have
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two times three, which is six. Cute.
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Simplify that five over seventy two. So these things
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, they're getting much smaller. Let's do another one
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a four after you simplify this rob even smaller and
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then a five is even smaller than that. So
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by looking at it, my guess is that the
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limit of a N equals zero. So let's go
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ahead and prove it, right? So first of
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all, I noticed that we have zero less than
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or equal to an because we have positive over a
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positive. Now we can write. This is one
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, three, five, and then go all the
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way up to two and minus one. Remember,
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Here we have n these air end different positions because
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these were the first and on imagers. And then
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instead of writing to end to the end power,
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we could just put to end write it out in
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times. Now, notice that each of these fractions
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is less than one. Therefore, this entire equation
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, this entire fraction is less than if we just
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go ahead and keep the one in the to end
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. Because here on then, in the middle,
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on the left side, all we're doing is multiplying
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one over to end by a number that's less than
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one. So if you get rid of these fractions
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, it's only going to make the remaining fraction bigger
40
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. So now, going on to the next page
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, we know that the limit of zero as n
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goes to infinity equals zero. We have the limit
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and goes to infinity one over two and boobs equal
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zero. Now, since we have zero less than
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or equal, say, and less than one over
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to end by the squeeze their own, which also
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holds for sequence. He sequences. We have that
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the limit yeah, of a N must equal this
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common value. The lower bound with zero the upper
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bond was zero. And because Anne's in between both
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of those sequences the limited and exists and also equal
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zero. So this proves our guests, and that's
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the final answer.