# Prove that two sequences converge to the same limit. [duplicate]

This question already has an answer here:

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|

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, YiFanMar 15 at 3:25

• $(\sqrt{a_n}+\sqrt{b_n})^2=a_n+b_n+2\sqrt{a_nb_n} \geq 0$ From here we get that $\frac{a_n+b_n}{2}+\sqrt{a_nb_n} \geq 0$ – 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 $$a=b$$.

• 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! – Buk Lau Nov 23 '18 at 14:16
• 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. – Siong Thye Goh Nov 23 '18 at 14:18
• 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? – Buk Lau Nov 23 '18 at 14:21
• 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? – Siong Thye Goh Nov 23 '18 at 14:28
• imgur.com/a/TaaijH9 is this the way to do it? Or did I write nonsense? Sorry if so!! Thank you. – 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 $$n>1$$.

• 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! – Buk Lau Nov 23 '18 at 14:40
• As $x,y$ are both positive it is the standard AM-GM inequality. So you also can use the result in formal proof. – Offlaw Nov 23 '18 at 15:04
• 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! – 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 $$n\rightarrow\infty.$$

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