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Numerade Educator



Problem 80 Hard Difficulty

A sequence $ \left\{ a_n \right\} $ is given by $ a_1 = \sqrt 2, a_{n + 1} = \sqrt {2 + a_n}. $
(a) By induction or otherwise, show that $ \left\{ a_n \right\} $ is increasing and bounded above by 3. Apply the Monotonic Sequence Theorem to show that $ \lim_{n\to\infty} a_n $ exists.
(b) Find $ \lim_{n\to\infty} a_n. $


(a) $$
\lim _{n \rightarrow \infty} a_{n}=2
(b) $$


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Video Transcript

let's use induction in party to show that this sequence is increasing. So first she's stable. Reprove ing so is increasing. So you prove by induction you. So now we have our bass case. This is where we'd like to show that a one is less than or equal to eight to. So we have a two equals radicals who plays they want using the formula. So that's just radical. Two plus room, too. And this is definitely bigger than room two, because here were adding something. So in other words, thiss thing right here it's bigger than route, too. If X is bigger than zero because we noticed the square, a graph is increasing, so the larger the input, the larger the square root. So this checks the bass case for us. Now we do the inductive step second part of the induction, so we'd like to go ahead and suppose a and minus one is less than or equal to K N. Now we want to show you that this inequality is true. After you increased and buy one on both sides there. So let's go ahead and write this on the next sage. This was given and recall the formula for Anne. So we would like to show so here, when we replace and within plus one recall the one that we wanted to show. So keep this one in mind. This is what we want Now this right inside we could expand this. This is just radical two plus an. So let me write this. And plus one equals this just by using this formula here, this hole's for all in. So just using that there that justifies list, then this is bigger than or equal to radical two plus and minus one. And this is due to this fact over here that's all have a hero indicating Oh, and then this is equal to an And that's exactly what we wanted a n plus one bigger than or equal to K and a so by induction we've just shown that the sequence is increasing. Now that that takes care of the first part of party. Now, we also we have another part, and this a and this is the show that a M is less than or equal to three. So part two were still in party here. Now, the second claim is that a n is less than or equal to three for all in will also prove this by induction. So that's the claim of here. Now we do proof by induction. Very similar last time in the meaning. What I mean by that is you still have your bass case. So this is where we want. We want this to be true for anyone now, is that true? Ai one, remember is radical too. And that is definitely less thinner, equal to three. So that's true. Now we go to the inductive step. This is where you suppose a M is less than or equal to three. And then we want to show that this is also true and you increase and buy one. So if we increase and buy one while for use the formal of recall the formula in the previous two ages, that's the formula for the ends. However, since AM is less than or equal to three inside the radical. If I just replaced the and with the three that gives me an upper bound here. So this fact justifies this inequality. This is radical five Bushes less than radical nine, which equals three. So we've just shown that inequality is this. Inequality here is true after you increase and buy one. So by induction there by induction, Yeah, so let's go to the next page. So we're last Cent last thing to say in part. Is that why it converges? So since a n you sees me is increasing, so in particular it's with its monotone. It's increasing and bounded. This was always just shown. It converges by the Monets on sequence there, so that completes party. Party is very short. So we're almost there. So we know that the ai converges so form part, eh? This last part in part A. We know at the limit of Ann. This is important. Let's call it l exist. It's a real number so say in our set of real numbers. So if we take this equation here and plus one equals radical two plus an, let's take the limit on both sides as n goes to infinity that here noticed that were increasing and buy one. But this is not going to change the limit because it's and goes to infinity and plus one also those to infinity. So we take the limit on both sides here on the left we have held on the right hand side we have inside the radical to plus l. So I just go ahead and take the limit on both sides. The left side becomes l on the right hand side. We could use the fact that the square root is continues to push the limit inside the radical and then the and goes to hell. Now licious do some algebra to solve this frail square. Both sides get that quadratic and then going to the next page, we can factor. This is L minus two l plus one people zero. This means that l equals two or l equals minus one. But we could actually eliminate this case over here. And the reason for this is that a n plus one equals radical two plus an is always bigger than Equal tau zero. Because the square room is always positive. You could recall the graph looks something like that. The smallest and I'LL ever be a zero here at the origin. Otherwise, it's positive, so the limit cannot possibly be negative on. Therefore, the only answer the limit, the one unique limit is two. And that's our final answer