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(a) If $ \left \{ a_n \right\} $ is convergent, show that

$ \displaystyle\lim_{n\to\infty} a_{n+1} = \displaystyle\lim_{n\to\infty} a_n $

(b) A sequence $ \left\{ a_n \right\} $ is defined by $ a_1 = 1 $ and $ a_{n + 1} = 1/(1 + a_n) $ for $ n \ge 1. $ Assuming that $ \left\{ a_n \right\} $ is convergent, find its limit.

(a) $n+1>N \rightarrow\left|a_{n+1}-s\right|<\varepsilon$

(b) $s=\frac{1+\sqrt{5}}{2}$

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Oregon State University

Harvey Mudd College

University of Michigan - Ann Arbor

So for part A, we're supposing that the sequence is conversion. So let's just go ahead and denote the limit by some letter s. That's the limit of the sequence. Now we'd like to show that if we add one to the end that this does that this new sequence also conversions to the same answer. So we'LL use the fact that by definition too and this section we can say for each Absalon bigger than zero there exist and capital and such that if we take the little end to be bigger than big and then a A and minus s is less than epsilon. So here we have n plus one self and is bigger than end then and plus one is also bigger than end. And this implies a N plus one minus s is lesson Absalon so basically down here. But we can go ahead and write that more formally. But that's basically the idea that will use here. So sense and plus one is bigger than end and little and his bigger than capital and by transitive ity of inequality and plus one is bigger than capital. And so by this condition, appear the distance between and plus one and s must be less than Absalon. So this shows that the limit of a N plus one is also equal to us. And this verifies part ay, that if you add one to the end, it does not change the limit. Now let's go ahead on to the next page for part B. So we're given a one equals one and you're even given a formula for an plus one. Assuming that it converges, Let's find the limit. So let's go on to the next page. So here and his bigger they're equal toe one. And we're supposing that the limit of a n exist. Let's just go ahead and call it s again, as we did before. So using part A. So by part, eh, the limit of a and plus one is also equal to us. So if we go ahead and take the limit of both sides one over one plus a n. So we know this equation. This is given. All we're doing is taking this equation and implying the limit on both sides. On the left hand side from part A. We know that limit is s and on the right hand side. This is just one over one plus s because the limit on Lee applies to an not the ones. Now let's take this equation here, multiply out the denominator to both sides. Then you get a square plus s minus one equals zero. Go ahead and use the contract IQ formula here, and you get two answers, and I claim that only one of these answers will work. You have a positive answer. You had a negative answer, and the negative answer is not allowed. And the reason is the following a one is one and we see it by our Rikers in formula, how to get the next term and a two and so on. So since we're starting a one equals one, this is positive. And then if we take a positive number and add one to it and then flip it over, this is still positive. So there's no way that the limit of a N could be positive. So we go to the next page here. So we have a one is bigger than zero. If we suppose a N is bigger than zero than a M plus one equals one over one plus a N is also bigger than zero. So a N is bigger than their equals. Zero Excuse me. Let me just say bigger than zero for all. And by mathematical induction. That's what I just showed above. The first term is one. It's positive if you suppose that it's true for and then also show that is true for M plus one. And that's what we did. Therefore, the limit of a N which equals s must also be bigger than or equal to zero. If you take unlimited positive numbers, though, worst case scenario is zero could never be negative. So this eliminates s equals one minus Route five over two. So we must have s equals one, plus Route five all over too. And that's our final answer.