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JH
Numerade Educator

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Problem 81 Hard Difficulty

Show that the sequence defined by
$ a_1 = 1 $
$ a_{n + 1} = 3 - \frac{1}{a_n} $
is increasing and $ a_n < 3 $ for all $ n. $ Deduce that $ \{ a_n \} $ is convergent and find its limit.

Answer

$L=\frac{3+\sqrt{3}}{2}$

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Video Transcript

we are given a one equals one first, let's show a N is always bigger than equals one for all in. So to do this, let's use induction. We have our bass case, which is an equals one. So we have a one equals one that's given up here. So that's bigger than our equals one. So that's a true statement. Now we go to the inductive step. This is where we suppose a N is bigger than equal tow. Want for some in so little and could be any number Now we'd like to show I want to show if we increase and buy one. This is the inductive step that it's still bigger than or equal to one. So to show this, let's just use the formula for and plus one that's given above circled in red three minus one over a m. However, what do we know about am that this is bigger than or equal to one? So in this case, this is bigger than our people, too. Three minus one over one because the most we could subtract is when the denominator smallest that denominator smallest when he had his one. This equals two and that is bigger than or equal to one. So by induction we have a n bigger than or equal to one. So let's keep that in mind and go on to the next page. So I'll just record that up here. We just showed a m bigger than or equal to one. Now let's show I am. The sequence is monotone increasing. Yeah, So here we list you also use induction areas for technique. We have our base case again. So here we like to show a one is less than a two. So a one from the previous page was equal to one A to use the formula. So three minus one equals two. So we see that, Yes, this is a true statement to is bigger than one. So the bass case holds. Now we go to the inductive step. So here we will suppose here's our inductive offices for some in so and could be any number, but it's just one particular fixed value. Now we do you want to use this to show I want to show that if you increase and buy one, so hear this end becomes and plus one and then on the right inside. Add one more to this end and add those ones together. This is what we'd like to show. So let's go ahead and do this. And plus two, three minus one over and plus one. Now we can go ahead and right the following and I'Ll explain why The reason we can't do this is by our Apophis is so if you just go ahead and take this equation, this inequality and blue if you go ahead and do some all's well, we can rewrite this and then multiple supply both sides by negative. So this means that if you have the negative, if you have the plus one on the end, it's a larger fraction. So here we were pleased a lot of refraction with the smaller fraction. So due to this inequality right here, this justifies this. And now, by using the formula on the previous page, we know this is equal to and plus one. So we started with employees too. We have bigger than people too, and then am plus one. So this proves by induction. So by induction, we conclude at the sequence AM is monotone increasing. So that's our second fact. We've used induction twice here. We just showed it was increasing. But on the first and second pages, we also showed that a and was bigger than ableto one. So we use this fact to show monotone increasing and we'Ll also use this fact again to show that the last thing we need to prove here before we find the limit is end less than or equal to three Alesha show. Excuse me Less than three. So we'll also use induction here our final induction his case and equals one. So here we had a one that was equal to one and that's definitely less than three. So the bass case, this is truth. Now let's go to the inductive step. So here, this is when we suppose that it's true for some end So we suppose inductive hypothesis for some men now increase and buy one. What We want to show that if you increase and buy one that it's still less than three So let's go ahead and increase and buy one. Then we use the formula given on page one. I now recall that we showed using this is where we use our first proof and is bigger than one. So that means that this fracture right here, this is a positive number. So we're subtracting a positive from three. So that's definitely less than three. So by induction, A M is less than three for all in and must go in and summarize what we have before we do the final step. We have that the sequence and is monotone increasing, and one is less than or equal to k n less than three for all end. So this was we proved each of these inequality separately, that both of those required induction. And we always also use induction to show increasing. So, Ron, we're almost done here. Let's say that serum by the mama tone sequence their own the limit of AM exist. So the sequences monotone and due to these inequalities over here, we see that and is bounded. And the monitoring sequence there, um says that if you have a monotone founded sequence than the sequence converges, or, in other words, the limit exists. So let's just go ahead and call it denoted by l. Oops. You okay? So here, if you increase and buy one and plus one still goes to infinity, so This is also L. So if you just take this three minus one over a end, take the limit on both sides. So on the left and then all soups also on the right. So on the left we get l three's constant. Take the limit inside the fraction you get one over l and then go ahead and multiply on bail. You get a quadratic use your quadratic formula here. So in this case, you should get two roots for this quadratic. He just used the formula. Now I recall that we had I am less than three baby within her equal Say one. Therefore, when we take the limit, the l should be inside the interval in maybe three. But in general, because you can't include the end points in general. So here, any l that satisfies this is that he is a possible answer. However, this number over here does not satisfy this. And so we are on ly left over with this value of our so that is the limit of the sequence. And that's the final answer