# Arithmetic-geometric mean convergence proof.

So in my book A friendly introduction to Analysis, there's an exercise that I'm having trouble with.

The exercise is as follow:

Consider sequences {$a_n$} and {$b_n$} which satisfy:
$0 < b_1 < a_1, a_{n+1}=\frac{a_n+b_n}{2},$ and $b_{n+1}=\sqrt{a_nb_n}$

(a) Show that $b_n<b_{n+1}<a_{n+1}<a_n$
I did this using an induction hypothesis that says $a_n<b_n$.

(b) Show that $0<a_{n+1}-b_{n+1}<\frac{a_1-b_1}{2^n}$
Now I can't quite solve this part. I can prove that: $a_{n+1}-b_{n+1}=\frac{a_n+b_n-2\sqrt{a_nb_n}}{2}<\frac{2a_n-2b_n}{2}=a_n-b_n$. But I don't know if this even relates to the problem. It just tells me the difference is decreasing.

(c) Deduce that {$a_n$} and {$b_n$} converge to the same value.
Even if I knew (a) and (b) were true I wouldn't know how to draw that conclusion from there. Perhaps if I could show that the difference of (b) grows infinitely small? But I wouldn't know how to do that.

A few exercises later I think there's an exercise which is related to this:
Prove that the following statements are equivalent:
(a) Completeness axiom: every nonempty subset $S$ of the real numbers that is bounded above has a least upper bound,
(b) Every monotone sequence that is bounded must converge.

I can easily proof that a implies b, since any monotone sequence that is bounded is one particular subset $S$. However I think the idea of that previous exercise should somehow be calculated into the proof that b implies a.

• Try substituting $b_1<a_1$ into the definitions of $a_n$ and $b_n$ – Simply Beautiful Art Nov 19 '16 at 18:17
• Do you mean to make an explicit formula for $a_n$ and $b_n$? – QuestionMaker Nov 19 '16 at 18:18
• – carmichael561 Nov 19 '16 at 18:23
• – Arnaud D. Mar 14 at 13:00

Use $-\sqrt{a_nb_n}\le -2b_n$ in \begin{align} a_{n+1}-b_{n+1} =\frac{a_n+b_n-2\sqrt{a_nb_n}}{2} =\frac{a_n+b_n-2b_n}{2} =\frac{a_n-b_n}{2} \end{align} and with induction the result of 2.) follows.

$0\le(\sqrt{a_n}-\sqrt{b_n})^2$ implies $$\sqrt{a_nb_n}\le\frac{a_n+b_n}{2}$$ while $b_n<a_n$ implies $b_n^2<a_nb_n$ and $a_n+b_n<2a_n$, which finishes 1.)

Hint

By induction, we prove that

$\forall n>0$

$$b_1\leq b_n\leq b_{n+1}\leq a_{n+1}\leq a_n \leq a_1$$

thus

$(a_n)_n$ is convergent as decreasing and bounded.

$(b_n)_n$ is convergent as increasing and bounded.

let $$l_a=\lim_{n\to+\infty}a_n$$

and

$$l_b=\lim_{n\to+\infty}b_n.$$

as we have

$$a_{n+1}=\frac{a_n+b_n}{2},$$

by passage to the limit, we get

$$l_a=\frac{l_a+l_b}{2}$$

and $l_a=l_b$.

• This is good mathematical argument and I've upvoted it (mainly because I love the AGM algorithm), but part (b) of the Question is one that the OP especially wanted to understand. Would you consider adding something about this to your Answer? – hardmath Nov 19 '16 at 18:36