# Solving a problem with Monotonic convergence.

1. Aug 19, 2008

### Disowned

1. The problem statement, all variables and given/known data

I have a problem that states that An = $$\sqrt{A_{n-1}}$$
, N is $$\geq$$ 1 where A0 is a positive constant.

I have to show that the sequence

{An} is decreasing if A0 > 1 and increasing if A0 < 1.

Show that the sequence is bounded below if A0 > 1 and bounded above
if A0 < 1.

Also I have to show the limit as n $$\rightarrow\infty A_n = 1$$

3. The attempt at a solution

Now I'm pretty sure that I am right so I just need some verifcation here. This is what I did:

A0 = $$\sqrt{A_{1-1}}$$ = $$\sqrt{A_0}$$

$$\sqrt{A_0}$$ < A_0[/tex]
A0 < (A0)2

Call A0 = F(n)
F(n) > 1
F(n) = $$\sqrt{2}$$. F(n) < N [$$\sqrt{2}$$ < 2]
F(n) < 1
F(n) = $$\sqrt{1/4}$$. F(n) > N [$$\sqrt{1/4}$$ = 1/2 > 1/4]

So {An} decreases if A0 > 1 and increases if A0 < 1.

Because of the behavior of An in relation to N, An will always be less than N itself. If An > 1 and will be bounded below N.

Likewise if An An < 1 An will be greater than N and will always be bounded above.

3.

n $$\stackrel{Lim}{\rightarrow}$$$$\infty$$ |$$\sqrt{A_n/A{n-1}}$$ = 1

Via the ratio test.

So, is this a valid approach or am I off?
1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

Last edited by a moderator: Aug 19, 2008
2. Aug 19, 2008

### HallsofIvy

You mean A1 is equal to $\sqrt{A_0}$

$$\sqrt{A_0}$$ < A_0[/tex][/quote]
This is only true if A0> 1. There are two different problems here, A> 1 and A< 1, and you need to distinguish between them.

I presume you mean An[/sup]= F(n). Other wise there is no "n" in F(n).

If you are dealing with the "A0> 1" case, F(0)> 1 and you need to prove this for other n.

What? Even assuming you mean A0, you are not given this!

Okay, I assume you are now doing the other case.

You have not proved that. Do so.

I would recommend using "proof by induction".

3. Aug 19, 2008

### Disowned

Okay based on what you are saying, If I rewrite the first part of the problem like this

A1 = $$\sqrt{A0}$$
And then write that this infers that A0 > A1
I could then say A0 > 1 ?

Then use that inequality to prove N > 1 by comparing An-1 > An?
Saying n -1>n would lead to n > n +1.

From there I guess using An = F(n) would be okay?