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We have two sequences: $$a_{n+1}=\sqrt{a_nb_n}$$ $$b_{n+1}=\frac{a_n+b_n}{2}$$ I need to prove that those are making Cantor's Lemma.(At the end I shold get that: $\lim_{n\to \infty}a_n=\lim_{n\to \infty}b_n$ by Cantor's Lemma)

Any ideas how?

Thank you.


marked as duplicate by Arnaud D., Gibbs, José Carlos Santos calculus Nov 12 '18 at 14:01

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  • $\begingroup$ Have you calculated the first few values of $a_n$ and $b_n$ for some starting values $a_1$ and $b_1$? Do you notice anything useful? $\endgroup$ – Steve Kass Nov 27 '13 at 22:22
  • $\begingroup$ I tried, but I didn't know where to start... $\endgroup$ – CS1 Nov 28 '13 at 6:58

If $a_1=b_1$, the two sequences are constant, and the proof is trivial. Otherwise, suppose $a_1<b_1$ or $b_1<a_1$. We have the property that for any two numbers $a<b$, $$a<GM<AM<b,$$ where $GM$ is the geometric mean, and $AM$ is the arithmetic mean. So, we must have $$a_1<a_2<b_2<b_1$$ or $$b_1<a_2<b_2<a_1.$$ In any case, for $n>1$ $$a_n<a_{n+1}<b_{n+1}<b_n$$ That shows us that $a_n$ is increasing and $b_n$ is decreasing. To apply the Cantor's lemma we must only show that $b_n-a_n\rightarrow 0$. To show that, we realize that $b_{n+1}-a_{n+1}<\frac{b_n-a_n}{2}$, since $b_{n+1}$ is at the middle of the points $a_n$ and $b_n$.

  • $\begingroup$ How do you prove it? that $a<GM<AM<b$? I don't see a way..... $\endgroup$ – CS1 Nov 28 '13 at 7:29
  • $\begingroup$ I can prove it bu induction? $\endgroup$ – CS1 Nov 28 '13 at 14:43
  • $\begingroup$ @YoavFridman No, you can't. Do you know what arithmetic mean is? Can you see why $\frac{a+b}{2}$ is in the middle of the interval $[a,b]$ on the real line? If yes, you should be able to prove $a\leq AM \leq b$. I explained the inequality $AM\leq GM$ under my answer. And the inequality $a\leq GM \leq b$ is obvious if you look at the squares. Alternatively, you can look at logarithms and then geometric mean becomes arithmetic mean. $\endgroup$ – savick01 Nov 28 '13 at 18:05
  • $\begingroup$ As savick01 said, you can prove it with the inequality $(\sqrt b-\sqrt a)^2\geq0$, and the inequality is strict if $b>a$. $\endgroup$ – Mateus Sampaio Nov 29 '13 at 12:57
  • $\begingroup$ You can prove the general inequality $AM\geq GM$ for $n$ positive numbers by induction, but is not as simple as many induction proofs. You can see it here: en.wikipedia.org/wiki/… $\endgroup$ – Mateus Sampaio Nov 29 '13 at 13:02

Hint. All you need is:

  • $\sqrt{ab} \leq \frac{a+b}{2}$ -- for the inequality between the two sequences
  • the distance between $\frac{a+b}{2}$ and $\sqrt{ab}$ is at most half of the distance between $a$ and $b$ because $\frac{a+b}{2}$ is just in the middle of the interval with ends $a$, $b$ and $\sqrt{ab}$ is somewhere inside it -- to show that $a_n-b_n\to 0$.
  • $\begingroup$ How the first one helps me? And I don't know how to prove those two things.. If you can help me with this it will be great... $\endgroup$ – CS1 Nov 28 '13 at 7:00
  • $\begingroup$ @YoavFridman The first one says that $GM\leq AM$. I skipped the fact that for $a\leq b$ we have $a\leq M\leq b$ where M is $AM$ or $GM$ as I considered it obvious. How to prove the first? Take squares of both sides of the inequality and do some calculations (you should get $(a-b)^2$ and notice that it is positive). What is unclear about the second? $\endgroup$ – savick01 Nov 28 '13 at 13:44
  • $\begingroup$ I didn't understand how you prove the first one. And I didn't understand how the second one helps me... Than k you! $\endgroup$ – CS1 Nov 28 '13 at 14:41
  • $\begingroup$ @YoavFridman Oh, c'mon, have you tried solving the first with the guidelines from my comment? There are like 4 steps and I gave you two of them and the other two are very simple. About the second: do you know what is the limit of $(\frac{1}{2})^n$? $\endgroup$ – savick01 Nov 28 '13 at 14:47
  • $\begingroup$ I'll try. I hope I wont stuck, else I'll ask. About the second is 0, right? $\endgroup$ – CS1 Nov 28 '13 at 22:00

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