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Given $a,b>0$ let $\{a_n\}$ and $\{b_n\}$ be sequences defined as follows: $a_1=a, b_1=b,a_{n+1}=\frac{a_n+b_n}{2},b_{n+1}=\sqrt{a_nb_n}$

Prove that the sequences converge and that their limits are equal.

I don't know how to begin to solve this question because it's the first time I encounter with a sequence defined by another inductive sequence. When I see an inductive sequence the tool I use is to show that the sequence is monotonic increasing/decreasing and that it's bounded and then use limit arithmetic to calculate the limit.

Thank you very much for your time and help.

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up vote 3 down vote accepted

First, it’s clear that if $a=b$, both sequences are constant, so without loss of generality let’s assume that $a<b$. Now $a_2$ is the arithmetic mean of $a_1$ and $b_1$, and $b_2$ is the geometric mean of $a_1$ and $b_1$, so $a_1<b_2<a_2<b_1$. Similarly, $a_3$ and $b_3$ are the arithmetic and geometric means of $a_2$ and $b_2$, so $b_2<b_3<a_3<a_2$. You have the tools to take it from here.

If you’re not familiar with the relationship between the arithmetic and geometric means, you may want to look here.

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given the inequality $\frac{a_n+b_n}{2} > \sqrt{a_nb_n}$ how do I prove that $a_1<b_2<a_2<b_1$? –  Anonymous Apr 9 '12 at 0:18
    
@Anonymous: You know that both the arithmetic and geometric mean of $a$ and $b$ lie between $a$ and $b$, and you know that the arithmetic mean is the larger of the two; what more do you need? –  Brian M. Scott Apr 9 '12 at 0:21
    
do I need to show now that b is the upper bound of both of the sequences and that the sequences are monotonically increasing for n>1? –  Anonymous Apr 9 '12 at 1:19
    
@Anonymous: Only one of them is. Write out a few more steps; you should find that $a_1<b_2<b_3<b_4\dots<a_4<a_3<a_2<b_1$. Thus, the $b_n$ are increasing for $n>1$ and bounded above by $b$, but the $a_n$ are ... ? –  Brian M. Scott Apr 9 '12 at 1:38
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@Anonymous: Now you’ve got it. And you should be able to use properties of limits to show that the two sequences actually converge to the same thing. –  Brian M. Scott Apr 9 '12 at 2:07
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