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Use a graph of the sequence to decide whether the sequence is convergent or divergent. If the sequence is convergent, guess the value of the limit from the graph and then prove your guess. (See the margin note on page 699 for advice on graphing sequence .)

$ a_n = \frac {1 \cdot 3 \cdot 5 \cdot \cdot \cdot \cdot \cdot (2n - 1)}{(2n)^n} $

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Harvey Mudd College

University of Michigan - Ann Arbor

University of Nottingham

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