Solving a problem with Monotonic convergence.

[Disowned]

Homework Statement

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

I have to show that the sequence

{A_n} 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

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 {A_n} decreases if A₀ > 1 and increases if A₀ < 1.

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

Likewise if A_n A_n < 1 A_n 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?

 

[HallsofIvy]

Homework Statement

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

I have to show that the sequence

{A_n} 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

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}

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.

A₀ < (A₀)²

Call A₀ = F(n)

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

F(n) > 1

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

F(n) = \sqrt{2}.

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

F(n) < N [\sqrt{2} < 2]
F(n) < 1

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

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

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

You have not proved that. Do so.

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

Likewise if A_n A_n < 1 A_n 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?

I would recommend using "proof by induction".

 

[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 A^n-1 > Aⁿ?
Saying n -1>n would lead to n > n +1.

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