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.

Thanks in advance.


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.)



By induction, we prove that

$\forall n>0$

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


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

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

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



as we have


by passage to the limit, we get


and $l_a=l_b$.

  • 2
    $\begingroup$ 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? $\endgroup$ – hardmath Nov 19 '16 at 18:36

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.