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I encountered this question in my homework:

$$a_1=x, b_1=y, \\ a_{n+1} =\frac{a_n+b_n}{2}, b_{n+1}= \sqrt{a_nb_n}, n\in \mathbb{N}$$

Given $x,y$ positive constants. I have to prove that they both converge to the same limit $L$.

I know that in order to prove that a sequence converges, it has to be (for example) Monotonically increasing and that $|a_n|<K$

I'm having difficulties proving with induction that either of the sequences is monotonic... how do I do so when the first sequence depends on the other? I know that by using the average inequality I get that $a_{n+1} > b_{n+1}$. How do I continue from here?

Thank you for your time and help!


marked as duplicate by Arnaud D., Eevee Trainer, Lee David Chung Lin, Song, YiFan Mar 15 at 3:25

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  • $\begingroup$ $(\sqrt{a_n}+\sqrt{b_n})^2=a_n+b_n+2\sqrt{a_nb_n} \geq 0$ From here we get that $\frac{a_n+b_n}{2}+\sqrt{a_nb_n} \geq 0$ $\endgroup$ – mathpadawan Nov 23 '18 at 14:00

By AM-GM inequality,

$$b_{n+1} \le a_{n+1}$$

Also, if $ b_n \le a_n$, then $a_{n+1} \le a_n$ since $a_{n+1}$ is the arithmetic means between $a_n$ and $b_n$.

Also if $b_n \le a_n$, then $b_{n+1} \ge b_n$ since $b_{n+1}$ is the geometric means between $a_n$ and $b_n$.

Hence we have $$b_{n+1} \le b_{n+2} \le a_{n+2} \le a_{n+1}$$

Both sequence converges. From $$ a=\frac{a+b}2$$

we can deduce that $a=b$.

  • $\begingroup$ Does that hold for any n? Or do I have to do it through induction? Do I have to assume that x>y or the other way around? Thank you! $\endgroup$ – Buk Lau Nov 23 '18 at 14:16
  • $\begingroup$ try to prove by induction from $n \ge 2$ onwards. we do not need assumption of $x>y$. it always hold for any positive pairs. $\endgroup$ – Siong Thye Goh Nov 23 '18 at 14:18
  • $\begingroup$ I never encountered such a sequence that relies on another sequencem I don't know how to even start, like how do I do it just for an for example? how do I take bn out of the picture? $\endgroup$ – Buk Lau Nov 23 '18 at 14:21
  • $\begingroup$ perhaps go slow, first check that $a_n$ and $b_n$ are positive. After that check that $a_{n+1} < b_{n+1}$. after that then prove the monotonicity. ping if you are stuck and update me which step are u stuck at? $\endgroup$ – Siong Thye Goh Nov 23 '18 at 14:28
  • $\begingroup$ imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you. $\endgroup$ – Buk Lau Nov 23 '18 at 15:01

if $x>y$, first observe that $x \ge a_n\ge a_{n+1} \ge b_{n+1} \ge b_n \ge y$ for all $n\in\mathbb{N}$

So both $\{a_n\}$ and $\{b_n\}$ converge. (as they are monotonic bounded)

Also $|a_n-b_n|\le|\frac{a_{n-1}+b_{n-1}}{2}-b_{n-1}|\le \frac{1}{2}|a_{n-1}-b_{n-1}|\le \frac{1}{2^{n-1}} |a_1-b_1| \to 0 $ as $n\to \infty$

So $\{a_n\}$ and $\{b_n\}$ converge at same limit.

if $y>x$ then $y \ge a_n\ge a_{n+1} \ge b_{n+1} \ge b_n \ge x$ for all $n>1$.

  • $\begingroup$ I did get to that inequity chain but I couldn't tell if it holds for all n, I just assumed it did, since all I could say for sure is that a2>a1 (when x>y). I know that it's true but do I have to show it for all n or can I just say it the way you did? Thank you! $\endgroup$ – Buk Lau Nov 23 '18 at 14:40
  • $\begingroup$ As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof. $\endgroup$ – Offlaw Nov 23 '18 at 15:04
  • $\begingroup$ This is exactly what I did but I thought that I still have to prove that it does hold for all n, I then got that they converge to the same limit the same way Siong Thye Goh mentioned in his answer. will this be enough as a formal proof? Thank you for your time! $\endgroup$ – Buk Lau Nov 23 '18 at 15:16

$|a_{n+1}-b_{n+1}|=\left|\frac{a_n+b_n}{2}-\sqrt{a_nb_n}\right|=1/2\left|a_n+b_n-2\sqrt{a_nb_n}\right|=1/2(\sqrt{a_n}-\sqrt{b_n})^2=(1/2)(1/4)(\sqrt{a_{n-1}}-\sqrt{b_{n-1}})^4=2^{-1-2}(\sqrt{a_{n-1}}-\sqrt{b_{n-1}})^{2\times 2}=\ldots=2^{-1-2-\ldots-n}(\sqrt{x}-\sqrt{y})^{2n}=2^{-n(n+1)/2}(\sqrt{x}-\sqrt{y})^{2n}\rightarrow 0$ as $n\rightarrow\infty.$

  • $\begingroup$ How did you get the third equality? $\endgroup$ – mathpadawan Nov 23 '18 at 14:09

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