Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

I am working on a multiple part question for an introductory Real Analysis course. I have part of it done, but I have some problems.

Let $0 < y_1 < x_1$, and set $$x_{n+1}=\frac{x_n +y_n}{2}, y_{n+1}=\sqrt{x_n y_n}$$ (a) Prove that $0<y_n <x_n$ for all $n \in \mathbb{N}$

For part (a) I believe I have proven $y_n < x_n$, but I am not sure if what I have is sufficient.

$$\sqrt{x_{n-1}y_{n-1}} < \frac{x_{n-1}+y_{n-1}}{2}$$

Rearranging yields

$$0<x_{n-1}+y_{n-1}-2\sqrt{x_{n-1}y_{n-1}}$$

Factoring,

$$0 < \left ( \sqrt{x_{n-1}} - \sqrt{y_{n-1}} \right )^{2}$$

So the above inequality should always hold, as long as $\sqrt{x_{n-1}} \neq \sqrt{y_{n-1}} \neq 0$, which I know to be the case for $n=2$.

(b) Prove that $y_n$ is increasing and bounded above, and that $x_n$ is decreasing and bounded below.

I have shown that $y_n$ is increasing. I need to know when $y_{n+1} > y_{n}$ for $n \in \mathbb{N}$. So,

$$\sqrt{x_n y_n}>y_n$$

This results in $x_n > y_n$. In other words, $y_n$ is increasing when it is less than $x_n$. I will have shown in part (a) that this is always true, and thus $y_n$ is monotone increasing. I have not been able to prove that $y_n$ is bounded above.

I have been able to prove that $x_n$ is decreasing, but not that it is bounded below. The inequality $x_{n+1} < x_n$ yields $y_n < x_n$, which I will proven to be true for all $n \in \mathbb{N}$.

(c) Prove that $x_{n+1} - y_{n+1} < \frac{x_1 - y_1}{2^n}$ for $n \in \mathbb{N}$

I know that $x_1 > y_1$ and $x_{n+1} > y_{n+1}$. I divide the first inequality by $2^n$ which yields

$$\frac{x_1}{2^n} > \frac{y_1}{2^n}$$

I've added the inequalities but clearly it does not help

$$y_{n+1} - x_{n+1} < \frac{x_1}{2^n} - \frac{y_1}{2^n}$$

(d) Show that $\lim_{n \to \infty} x_n = \lim_{n \to \infty} y_n$

Assuming I am able to prove that $x_n$ is decreasing and bounded below, and $y_n$ is increasing and bounded above, I can use the Monotone Convergence Theorem.

$$\lim_{n \to \infty} x_n = L$$ $$\lim_{n \to \infty} y_n = W$$

$$L = \lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} \left ( \frac{x_n}{2}+\frac{y_n}{2} \right )= 1/2 \cdot \lim_{n \to \infty} x_n + 1/2 \cdot \lim_{n \to \infty} y_n $$

Which yields $L=W$. Any help with what I've missed would be great.

share|improve this question
    
I have a quick side question. If a sequence $b_n$ is convergent, does $\lim_{n \to \infty} b_n = \lim_{n \to \infty} b_{n-1}$ ? –  Delightful_Richard Jan 25 '13 at 12:44
add comment

1 Answer

up vote 0 down vote accepted

In principle, all st fine as you do. (a) is just the famous inequality betqeen artihmetic and geometric mean of positive numbers. In (b) you do the right calculations, but formulate the logic in the wrong direction. We have and need $x_n>y_n\Rightarrow \sqrt {x_ny_n}>y_n$, while you wrote it down the other way round. Once you have that $y_n$ is increasing and $x_n$ decreasing, you have $x_1$ as upper bound for $y_n$ and $y_1$ as lower bound for $x_n$: $y_1\le y_n<x_n\le x_1$. In (d) you might also use (knowing the existence of $L$ and $W$ that $|L-W|<\frac{x_1-y_1}{2^n}$ for arbitrary $n$, hence $L=W$.

For (c), make use of $$x_{n+1}-y_{n+1}\le \frac{x_n+y_n}2-y_{n+1}\le \frac{x_n+y_n}2-y_{n}=\frac{x_n-y_n}2,$$ which uses $y_{n+1}>y_n$ and then do induction.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.