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Solving a problem with Monotonic convergence.

  1. Aug 19, 2008 #1
    1. The problem statement, all variables and given/known data

    I have a problem that states that An = [tex]\sqrt{A_{n-1}}[/tex]
    , N is [tex]\geq[/tex] 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 [tex]\rightarrow\infty A_n = 1[/tex]




    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 = [tex]\sqrt{A_{1-1}}[/tex] = [tex]\sqrt{A_0}[/tex]

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

    Call A0 = F(n)
    F(n) > 1
    F(n) = [tex]\sqrt{2}[/tex]. F(n) < N [[tex]\sqrt{2}[/tex] < 2]
    F(n) < 1
    F(n) = [tex]\sqrt{1/4}[/tex]. F(n) > N [[tex]\sqrt{1/4}[/tex] = 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 [tex]\stackrel{Lim}{\rightarrow}[/tex][tex]\infty[/tex] |[tex]\sqrt{A_n/A{n-1}}[/tex] = 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. jcsd
  3. Aug 19, 2008 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You mean A1 is equal to [itex]\sqrt{A_0}[/itex]


    [tex]\sqrt{A_0}[/tex] < 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".
     
  4. Aug 19, 2008 #3
    Okay based on what you are saying, If I rewrite the first part of the problem like this

    A1 = [tex]\sqrt{A0}[/tex]
    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?
     
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