3
$\begingroup$

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.

$\endgroup$
3
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ given the inequality $\frac{a_n+b_n}{2} > \sqrt{a_nb_n}$ how do I prove that $a_1<b_2<a_2<b_1$? $\endgroup$ – Anonymous Apr 9 '12 at 0:18
  • $\begingroup$ @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? $\endgroup$ – Brian M. Scott Apr 9 '12 at 0:21
  • $\begingroup$ 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? $\endgroup$ – Anonymous Apr 9 '12 at 1:19
  • $\begingroup$ @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 ... ? $\endgroup$ – Brian M. Scott Apr 9 '12 at 1:38
  • 1
    $\begingroup$ @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. $\endgroup$ – Brian M. Scott Apr 9 '12 at 2:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.