Prove that two sequences converge to the same limit.

Question

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!

Answer 1

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

Answer 2

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 $a=b$.

Answer 3

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 $n>1$.