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Sequence: nondecreasing, bounded above, prove s_n < L

  1. Oct 6, 2009 #1
    1. The problem statement, all variables and given/known data
    If {s_n} is nondecreasing and bounded above, and L = lim s_n, prove that s_n <= L.


    2. Relevant equations



    3. The attempt at a solution
    This is one of those proofs that seems, to me, to be obvious from the proven theorem that states that the limit of a sequence is equal to the least upper bound. But I doubt my professor would both with it if it were obvious. Please help!
     
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  3. Oct 6, 2009 #2

    Dick

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    Mmm. Proof by contradiction? Assume there is an s_k such that s_k>L. Now what?
     
  4. Oct 6, 2009 #3
    If s_k > L, then L is not a least upper bound for s_k and, hence, L is not the limit for s_k. This is a contradiction, so L < s_k? How about that?
     
  5. Oct 6, 2009 #4

    Dick

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    I guess that's ok, if you know that a bounded nondecreasing sequence has a limit that is the LUB of the sequence. You could maybe be a little more direct and define s_k-L=e>0. Then since the sequence is nondecreasing, s_i-L>e for all i>k. L is not the limit.
     
    Last edited: Oct 6, 2009
  6. Oct 6, 2009 #5
    We did just prove a theorem that the least upper bound = the limit of a nondecreasing, convergent sequence, but... it does seem like we always come back to the epsilon proof. Along that line:

    If e > 0, we need to find N such that s_k - L < e. But since s_k is nondecreasing, there is i > 0 such that s_i - L >= e, so L is not the limit. But this contradicts our hypothesis, so in fact, s_n < L. End.

    This does make sense to me, although I wouldn't have come up with a different index, i. e., "i", unless you had shown me that.
     
  7. Oct 6, 2009 #6

    Dick

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    You've got the idea, but the use of the indices is still sloppy. If there were AN s_k>L and e=s_k-L then s_i-L>=s_k-L=e for ALL i>k. If the lim(s)=L then there is AN N such that |s_i-L|<e for ALL i>N. Clearly the two conflict for i>max(N,k). Be sure and distinguish between indices like 'k' and 'N' which identify a single element of the sequence and indices like 'i' which hold for a whole range.
     
  8. Oct 6, 2009 #7
    I am very confused by all the different indices, but I have gotten this far: Assume there is an s_k such that s_k > L. Then for e > 0, s_k - L = e. But since the sequence is nondecreasing, there is also an s_i - L > s_k - L = e. If L is the limit, then there exists an N such that |s_i - L| < e for all i >=N. But

    I can't quite make the conclusion. It seems to me as if finding s_i - L > s_k - L should be a conclusion, but that doesn't seem quite complete. I really appreciate your help.
     
  9. Oct 6, 2009 #8
    And, hitting the middle between my original thought and your suggestions, please look at this version:

    Suppose that there is an s_k such that s_k > L. Then for e e> 0, s_k - L >= e. But {s_n} is nondecreasing and bounded above, so s_n has a least upper bound, say M. By theorem ( a nondecreasing sequence which is bounded above is convergent, that is, L = M), lim s_n = M, so we have a contradiction. That is, if M is the least upper bound for s_n, then M = L, so that s_k cannot be greater than L. Therefore, s_k <= L implies that s_n <= L. End of proof.
     
  10. Oct 6, 2009 #9

    Dick

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    i) Don't just pick any epsilon. Pick e=s_k-L. ii) Since the sequence is nondecreasing s_i-L>=s_k-L=e FOR ALL i>k. A single value of i isn't enough. If you just had a single value of i such that s_i-L>=e, you might be able pick N bigger that that value of i. But if it's true for ALL i>k then there clearly is no such N.
     
  11. Oct 6, 2009 #10

    Dick

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    Actually, your original proof was fine. You are just making it more complicated without making it better. By your theorem you know the limit, L, is the LUB of the s_k. If s_k>L then L can't be the LUB. That's a contradiction, so s_k is not greater than L.
     
  12. Oct 6, 2009 #11
    Thanks so much! You cannot imagine how that helps my confidence. Have a great evening.
     
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