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Cauchy => bounded?

  1. Jun 9, 2007 #1
    1. The problem statement, all variables and given/known data
    Thm: If a sequence is Cauchy than that sequence is bounded.

    However Take the partial sums of the series (sigma,n->infinity)(1/n). The partial sums form a series which is Cauchy. But the series diverges so the sequence of partial sums is unbounded.

    Sequence of partial sums is Cauchy b/c d(1/x,1/(x+1))=1/(k(k+1)) -> 0 as k->infinity

    Have I done something wrong?
     
    Last edited: Jun 9, 2007
  2. jcsd
  3. Jun 9, 2007 #2

    matt grime

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    Why do you think the sequences of partial sums are cauchy? They aren't.

    I cannot begin to decipher what b/c d(..... means.

    If S_n means the sum to n terms, what makes you think that given e>0, there is an n(e) such that for all n,m>n(e) we have |S_n - S_m| < e?

    It is easy to show this is impossible, using the same idea as to show that the series itself diverges. Remember that

    1+1/2+1/3+1/4+1/5+... > 1+(1/2+1/2)+(1/4+1/4+1/4+1/4)+(1/8+1/8+1/8+...) > 1+1+1+1+...
     
    Last edited: Jun 9, 2007
  4. Jun 9, 2007 #3

    siddharth

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    Is b/c d(.., "because" , the "distance" ...?

    As matt suggested, I think you should be less cryptic in your notation.
     
  5. Jun 9, 2007 #4

    AKG

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    Yes, that's not the definition of Cauchy. A sequence a(n) is Cauchy if for all E > 0, there exist N natural such that for all m and n > N, |a(n) - a(m)| < E. All you've shown is that for all E > 0, there exists some natural N such that |a(N) - a(N+1)| < E, or something like that.
     
  6. Jun 9, 2007 #5
    oh... I thought it was obvious. I actually used b/c in my exam recently to indicate 'because'. Although I used it in the middle of a sentence so hopefully might be less confusing than here.
     
  7. Jun 10, 2007 #6

    matt grime

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    If you'd've put a 'The' at the start of the sentence to actually make it into a sentence that might have helped. Anyway, have you understood the explanation of your mistake?
     
  8. Jun 10, 2007 #7
    I think I have. Each term in the sequence is a series summed over a finite number of terms. However the larger the number of terms, the larger the sequence gets. Very much unlike Cauchy seqences where a recquirement is that each term in the seqence decreases in magnitude the further out the sequence gets.
     
  9. Jun 10, 2007 #8

    matt grime

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    it is perfectly possible to be strictly increasing and still bounded/cauchy


    nope, there is nothing that states the terms have to decrease in magnitude. It is the difference between terms in the sequence that is important.
     
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