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Limit of a sequence

  1. Sep 15, 2012 #1
    1. The problem statement, all variables and given/known data

    Show that if sn[itex]\leq[/itex]b for all but finitely many n, then lim sn[itex]\leq[/itex]b.




    2. Relevant equations



    3. The attempt at a solution

    My question is regarding the absolute value portion of the proof:

    by contradiction: Call lim sn s. Suppose s>b. Then

    l sn-s l < [itex]\epsilon[/itex].

    Choose [itex]\epsilon[/itex]=s-b.

    Then l sn-s l < s-b.

    -(sn-s)<s-b

    sn>b, contradiction to problem statement.



    My question is this: The proof only seems to work if we assume sn-s is negative. But why couldn't sn be greater than s? Am I missing something important in the wording of the problem? Perhaps the "all but finitely man n"? I actually don't quite grasp what that means.

    If the sequence was 1/n



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



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 16, 2012 #2

    SammyS

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    If l sn - s l < s - b ,

    then -(s - b) < sn - s < s - b .

    That's -s + b < sn - s < s - b .

    Add s to all : b < sn .

    As for the "all but finitely many n":

    That's going to be an important part of the proof.

    Certainly, of those values of n, for which sn > b, one of those n's is largest, call it N. What does that say about sn if n > N?
     
    Last edited: Sep 16, 2012
  4. Sep 16, 2012 #3
    Ok. I've looked over the problem for a while now and I understand the absolute value portion. You still considered when sn-s is positive but it didn't affect the contradiction.

    The only conclusion I am able to make about your last question, and I don't know how this affects the proof, is that s(n) doesn't exist where n>N.
     
  5. Sep 16, 2012 #4
    Wait, but that means there is no N in Naturals such that n>N implies ls(n)-sl<epsilon, for some epsilon >0. So that is the contradiction? That the limit doesn't exist?
     
  6. Sep 16, 2012 #5

    SammyS

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    Maybe I could have stated it better.

    There are only a finite number of n values for which sn > b . Right.

    Let N be the index (subscript) of the last sn for which sn ≥ b.

    I.e. if sn ≥ b, then n ≤ N.

    So, if n > N, then sn < b.
     
  7. Sep 16, 2012 #6

    Ray Vickson

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    Suppose r = s-b > 0. Choose any ε > 0, ε < r. There exists N so that for all n ≥ N we have
    |s-sn| < ε, meaning that s - ε < sn < s + ε, so sn > s-ε > s-r = b, and this contradicts the original hypothesis.

    RGV
     
  8. Sep 16, 2012 #7

    aha! I see it now! Confusion between notation of s(n) and n's was throwing me off in the wrong direction (literally, on the number line, in the wrong direction!).

    Thank you for your patience and guidance!
     
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