Question about arithmetic–geometric mean

Question

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.

Answer 1

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 $a_n-b_n\to 0$.

Answer 2

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 $b_n$.