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(a) Show that if $ \lim_{n \to \infty} a_{2n} = L $ and $ \lim_{n \to\infty} a_{2n + 1} = L, $ then $ \{ a_n \} $ is convergent and $ \lim_{n \to \infty} a_n = L $.

(b) If $ a_1 = 1 $ and

$ a_{n + 1} = 1 + \frac {1}{1 + a_n} $

find the first eight terms of the sequence $ \{ a_n \} $. Then use part (a) to show that $ \lim_{n \to \infty} a_n = \sqrt{ 2 } $. This gives the continued fraction expansion

$ \sqrt{ 2 } = 1 + \frac {1}{ 2 + \frac {1}{2 + \cdot \cdot \cdot}} $

(a) If $n$ is even, $n >n_{0} > 2 \max \left\{n_{1}, n_{2}\right\}+ 1 > 2 \max \left\{n_{1}, n_{2}\right\} \Rightarrow$

$\frac{n}{2} > \max \left\{n_{1}, n_{2}\right\} \Rightarrow \frac{n}{2} > n_{1} \Rightarrow\left|a_{2\left(\frac{n}{2}\right)}-L\right| < \epsilon \Rightarrow$

$\left|a_{n}-L\right|<\epsilon$

If $n$ is odd, $n>n_{0}>2 \max \left\{n_{1}, n_{2}\right\}+1 \Rightarrow n-$

$1>2 \max \left\{n_{1}, n_{2}\right\} \Rightarrow \frac{n-1}{2}>\max \left\{n_{1}, n_{2}\right\} \Rightarrow \frac{n-1}{2}>$

$n_{2} \Rightarrow\left|a_{2\left(\frac{n-1}{2}\right)+1}-L\right|<\epsilon \Rightarrow\left|a_{n}-L\right|<\epsilon$

(b) Consecutive terms approach the same number $L$ as $n \rightarrow \infty,$ so substitute $L$

into the formula for $a_{n+1}$ and solve for $L .$

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Campbell University

Harvey Mudd College

Baylor University

Boston College

in this problem were given that the sequence A to end and also the sequence a two and plus one, that the's both converged to the same limit. L So, by definition, let's go ahead and let absolutely a positive number. So by this fact here there exists some positive number. Let's call it and won such that if we go ahead and take and larger than this. And similarly, there's another Inger we can say on DH, too, such that when we take and the larger than this end, too, we get a similar inequality. But this time involving two in plus one. Now the next step is to choose little end large enough so that we can have both of these at the same time. So if we have both of these inequalities at the same time, if you notice here, this's all the even ends. And then this is all the odd values. And so every number and that we choose is going to be even or odd, and we'd like to say eventually this is what we inequality that we desire. So we'LL go ahead and let and we'LL take end to be really large. Let's take it to be larger than two times. Let me take the max the larger of and one and two. And let me add one to that. The reason I'm putting the the two the two times and then the plus one is because of. I want to make sure that if I multiply this end here by two and add one that I could use both of these inequalities over here. If you notice these have to end in two and plus one. Where is the one down here on Lee has an end. That's why I have to make up for it by including the two and the plus one just to ensure. So I'll just put here in parentheses insurers we can use and as we mentioned now, because now that we fixed little and to be larger than this number over here, further every value and we'll be even rods that means for each value and will be able to apply the first or second known inequality. Therefore, taking little end to be larger than the circle number over here implies the desire, inequality and that proofs therefore lim a and equals l. And that's what we saw out to prove that completes party. Now let's go ahead two part B. So we're given some information here. You're given the first term of the sequence and you're given the n plus one term recursive lee. So the first step here in part be used to just go ahead and compute some values here. So let's just list the values of end all the way up, Tio, and then here will have our a n. So a one here that was given to just be one now in general will have to use this formula. But let me just go ahead and and show it for a two. By definition. Here this is and I get that by plugging in and equals one into the formula. You plug that into the formula and then you arrive here and that becomes one point five. And similarly, if he wanted a three, you would do one plus one over one plus a two and then you would go ahead and plug in your previous value a two and so on and you'LL keep going there. So I'll just go ahead and fill out their remaining values here that comes one point for And then after this, the decimal start getting longer here. So I was just round off. No. And it looks like from our pattern here that the sequence looks like it's converging rather quickly to about one point four one four. So this looks like it converges, and we don't know exactly what the number is delicious. Call it El now from part, eh? We know that if we're now here, we're assuming Ellis Limit of a M. That's the limit of the sequence and from party A. We know that this sequence here, if you just add one to the end that's just the same is and so remember every sort, like if this and is even then n plus one is our similarly if and his are than endless one is even so. These terms are one of them's always the opposite of the other. In terms of even a rod and from party. We showed that they both converge to the same limit. So that means if we take the limit of both sides of this, let him go to infinity. Then the equation becomes l same limit one plus one over one plus and then dilemma dear. Also Ezel. And then here we can go ahead and multiply both sides by this helpless one. But and they're out this cancel ofsome those l's you could l squared is too. And at this point, we have plus or minus square root. But we can see here that we should choose a positive on because this is conversion toe a positive number. So swear word of two is the limit of the sequence, and that's the final answer.