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Let $ a $ and $ b $ be positive numbers with $ a > b. $ Let $ a_1 $ be their arithmetic mean and $ b_1 $ their geometric mean:

$ a_1 = \frac {a + b}{2} $

$ b_1 = \sqrt {ab} $

Repeat this process so that, in general,

$ a^{n + 1} = \frac {a_n + b_n}{2} $

$ b_{n + 1} = \sqrt {a_n b_n} $

(a) Use mathematical induction to show that

$ a_n > a_{n + 1} > b_{n + 1} > b_n $

(b) Deduce that both $ \{ a_n \} $ and $ \{ b_n \} $ are convergent.

(c) Show that $ \lim_{n \to\infty} a_n = \lim_{n \to \infty} b_n $. Gauss called the common value of these limits the arithmetic-geometric mean of the numbers $ a $ and $ b. $

a. it is true for all $n$ by mathematical induction.

b. So they are both convergent by the Monotonic Sequence Theorem.

c. \alpha=\beta

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Missouri State University

Campbell University

Harvey Mudd College

Baylor University

So here we're given some positive numbers and be with a larger than be Kim. And then we define some sequence a n and being by the following equations and for convenience, we could also just call a a subzero. So these will be the first terms of the sequence instead of a one and be one. So we'd like to go ahead and show this these inequalities here. So we go by induction. And first we have our bass case, which now we can call and equal zero, so we'd like to show. So let's go ahead and show these one at a time, please. So let's go ahead and deal with this one. First we have a one, by definition, is just a plus B over two. Now, since B is less than a, this is less than a plus a over to some using the given information over here, and that's just equal toe A or something is enough. So that guarantees this now similarly, for B one, go to the definition over here and now this is larger than this squared of B times being, because once again is larger than me. And this is just equal tto be or as we call it, being up that guarantees the second one Now for the third one instead of just dealing with a one and be one, let me actually show this is true for any end because well, actually need this in the inductive step is wells the bass case Soldiers go ahead instead of plugging in an equal zero I'LL use any end now This is what we want to show but this is equivalent to you're the definition of the Angelus won and be endless one These were given information Appear now this Now I'LL bring this inequality Let me running out of room here. So bring this over here Let's go ahead and multiply both sides by two and in square both sides some boiling it out here on the left. Oops! Plus being square and then the right side I get a four there and then the radical goes away and then go ahead and subtract this force of the other side And I could go ahead and factor the left hand side and that becomes a and minus the end and that's square bigger than zero. Now, in the bass case, that we're dealing with this becomes a one is bigger than be one if and only if eh not minus B not squared is bigger than zero. And this is true because a minus B not. It's just a minus B. And this is a positive number. By the given information. Have you square it? It's also going to be a positive number. So that guarantees this inequality here. But that's equivalent to what we were trying to show. And so that's the third inequality in the Bass case. I'm running out of room here, so I will need to go to the next page. However, Will will want to remember one. We're proving this inequality here in the middle. We just did it for the Bass case Now in the General case, also referred to this expression over here when we showed that the desire inequality was equivalent to this inequality here. So that's what we'LL want to memorize in a few moments. So now let me go on to the next page. But that does complete party. Excuse me? The thank you and sorry that completes the bass case of party. We still have to do the induction inductions, though, so suppose that the inequality is true for n equals K. So this means that the inequalities that I'm writing care of true, this's corresponding to an equal scan And then we want the show show its true for K plus one. So we're increasing okay by one there and that'll be the inductive part of the mathematical induction. So now we have I'm going to deal with this one first. Hey, K plus one, go to the definition again. Now, by our hypothesis, I know that this is less than because I know B k is less than a k so that lets me use the inequality here and then just simplifying this. I just get a K and that gives me the first inequality. Similarly, for the last one go to the definition of being plus one. Now this is bigger than D k. Time's the UK because once again, be case less than a day, and that is just equal to B. K. And that gives us the last inequality there. And now for the middle inequality going back to what we mentioned on the previous page, this was true if and only if a K minus V. K. Was positive the square was positive. But this will be true here because once again, be case less than a K. And so that gives us our desired inequality, which we derived on the previous page. And so that justifies the middle inequality there. And now we've used induction to complete party. So let's move on to a party. So we want to show that the sequences converse so from part ay, we already showed the following part of our inequalities were these two in particular. Note that this is saying that this means and is a decreasing sequence and this one says bn is increasing. So we're trying to set ourselves up to use monotone convergence there. However, we have monitor necessity here for Ann and being, but we also have to show they're bounded. So now we know that some party again and is larger than the end. That was the third inequality for apartment. But since being is decreasing, I can go along, keep using these inequalities here, and targets have B. So this shows that the end sequence is founded bounded below, so change color here. So for a and in particular you have that it's decreasing and bounded below now for being We see that it's increasing and we'll show that also that it's found it about so using the same inequality is above just switching the order here now is the fact that a N is freezing and we see that being is founded above. Therefore, by the Monets phone commercials serum both the sequences converge and that results for me going on to part see, will show the following that they actually converge to the same number. So we can assume this. This is what we want to show here. So I'm gonna just put this in parentheses. This's not part of our work here, So let me go ahead and call the limit of a N l one and let me call the limit of BNL too. So we're not supposing these air equal yet. However, we do know the following by definition of and sequence and I'LL go ahead and take and to go to infinity. So we're taking a limit on both sides and we know that this will converge. Teo Ella's well, so here you could be a little more formal here if you need to prove that he's converse with the same number. L So when you go to the side and say, Let Absalon big with zero there is this What's wrong symbol there? Sorry about that. There exists a number end such that if n is bigger than end A and minus one is less than absolute and sense. If n is bigger than end, then n plus one is also bigger than him. And that would imply so. This implies women of a and plus one is also equal toe l one. So now when I take the limit on both sides, I could write the left hand side is l and then the right side Oops l one and the right side becomes l one plus l two over too. And if we could just go ahead and simplify this a little bit, we could see l one equals lt. But it's a track l one from each side and you're left with l one equals l too therefore limited and equals limited bm. And that completes Parsi