# Arthur Mattuck, Introduction to Analysis, Problem 5-7

1. Oct 5, 2007

### EMR

This question is from Arthur Mattuck's "Introduction to Analysis", chapter 5, problem 5-7.

1. The problem statement, all variables and given/known data
Define a sequence recursively by $$a_{n+1}=\sqrt{2a_{n}}$$, $$a_{0}>0$$.

(a) Prove that for any choice of $$a_{0}>0$$, the sequence is monotone and bounded.

2. Relevant equations

None

3. The attempt at a solution

I've rewritten the sequence as

$$a_n = \sqrt{2{\sqrt{2\sqrt{2\sqrt{2...\sqrt{2a_0}}}}}$$,

and taken the ratio of $$a_{n+1}$$ and $$a_{n}$$, which leads me to the expression

$$\frac{a_{n+1}}{a_n} = 2^{(1/2)^{n+1}}\sqrt{a_0}$$.

From computing the sequence for a few initial values of $$a_0$$, I've been able to determine that the sequence is decreasing if $$a_0>2$$ and increasing if $$0<a_0<2$$. However, I'm not sure how to show this from this ratio, i.e. how does

$$2^{(1/2)^{n+1}}\sqrt{a_0}<1$$

which indicates that the sequence is decreasing, turn into $$a_0>2$$?

The rest of the problem comes easy enough. I'm able to show that it is bounded and that the limit is 2 (in part b of the question).

2. Oct 5, 2007

### Dick

You are making this too complicated. Proving a_(n+1)>a_n translates into sqrt(2*a_n)>a_n. For what range of x is it true that sqrt(2x)>x.