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Homework Help: Prove that this sequence converges

  1. Mar 22, 2007 #1
    1. The problem statement, all variables and given/known data

    Let [itex]a_{0} = a >1 [/itex] and let [itex]a_{n+1} = a^{a_n}[/itex].
    Show that {[itex] a_{n} [/itex]} comverges for [itex] a < e^{e^-1} = 1.4446678 [/itex]

    2. Relevant equations

    This is a Theorm I learned in Real Analysis and hope to apply it to this problem:

    Theorm: If a sequence is montonically increasing and bounded, then it is convergent

    3. The attempt at a solution

    {[itex] a_{n} [/itex]} = {[itex]a, a^a, a^{a^a}, ...[/itex]}

    Clearly, {[itex] a_{n} [/itex]} is monotonically increasing is is bounded below by a.

    How do I show that it is bounded above?
    Last edited by a moderator: Mar 22, 2007
  2. jcsd
  3. Mar 22, 2007 #2
    try induction assuming that a(o) is less than the maximum a value...
  4. Mar 22, 2007 #3

    Sorry, I dont understand what you are trying to say...


    [itex]a < a^a < a^{a^a} < ...[/itex] ... So I am having a hard time explaining how this is bounded for a>1
  5. Mar 22, 2007 #4
    This doesn't mean anything. A sequence could be increasing and still be bounded. And [tex]e^{e-1} = 5.574941539 [/tex]. I don't think that's what you meant....
    Last edited: Mar 22, 2007
  6. Mar 22, 2007 #5


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    He meant e^(1/e). The formatting didn't come out 100% clear.
  7. Mar 22, 2007 #6


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    If a<e^(1/e) and x<e, what can you tell me about a^x?
  8. Mar 22, 2007 #7
    Thank you for the correction Dick...I allways have trouble with this latex stuff :confused:

    If a<e^(1/e) and x<e, we can conclude that a^x < e

    so a^x < e



    a < e^(1/x) ...

    Can you please give me another push ?
  9. Mar 22, 2007 #8


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    Ok. a=a0<e^(1/e)<e. a1=a^a0. So a1<e. a2=a^a1. a1<e. So a2<e. a3=a^a2. a2<e. So a3<e. a4=a^a3. a3<e. So a4<e. How long do you want me to keep this up?
  10. Mar 23, 2007 #9
    Okay...I see it now.

    it is bounded above by e.


    for 1 < a < e^(1/e)
  11. Mar 23, 2007 #10


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    You've got it.
  12. Mar 24, 2007 #11
    I can follow your example, but when i practicly pund these numers in my calculator, i find that it is not bounded by e.

    for example,

    a = 1.3

    a < a < 1^(1/e) = 1.444....

    (1.3)^(1.3)^(1.3)^.....^(1.3) >>> e ... hence, not bounded by e.
  13. Mar 25, 2007 #12


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    (%o1) f(x) := 1.3^x;
    (%i2) f(1.3);
    (%o2) 1.406456673237886
    (%i3) f(%);
    (%o3) 1.446293346285982
    (%i4) f(%);
    (%o4) 1.461488869891772
    (%i5) f(%);
    (%o5) 1.467327108831111
    (%i6) f(%);
    (%o6) 1.469576402415039
    (%i7) f(%);
    (%o7) 1.470443905739509
    (%i8) f(%);
    (%o8) 1.470778619613915
    (%i9) f(%);
    (%o9) 1.470907784591691
    (%i10) f(%);
    (%o10) 1.470957631962913
    (%i11) f(%);
    (%o11) 1.470976869521044
    (%i12) f(%);
    (%o12) 1.470984293924533
    (%i13) f(%);
    (%o13) 1.470987159254924

    Are you sure the calculator isn't doing f(x):=x^1.3? That's quite different.
  14. Mar 26, 2007 #13
    >> a = 1.3

    a =


    >> a^a^a^a^a^a^a^a^a^a^a^a

    ans =


    >> (((((((a^a)^a)^a)^a)^a)^a)^a)

    ans =


    Perhaps i am miss-understanding how you are computing it.
  15. Mar 27, 2007 #14
    You are computing:
    [tex] a_n = (a_{n-1})^a[/tex]

    You need:
    [tex] a_n = a^{a_{n-1}} [/tex]

    It's a subtle difference in this case.
  16. Mar 27, 2007 #15


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    Which do you believe, your reasoning or your calculator?

    (%i8) a^(a^(a^(a^(a^(a^(a^a))))));
    (%o8) 1.470778619613915
  17. Mar 27, 2007 #16
    Why would my calculator/matlab give a different answer.

    The reasoing is correct, so shouldnt the calculator give the same answer as your showing?
  18. Mar 27, 2007 #17


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    Is the warranty still valid on the calculator? Seriously, did you explicitly parenthesize it like I did? Powers are not associative.
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