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Does my proof work?

  1. Sep 21, 2008 #1
    Does my "proof" work?

    n\in \mathbb{N}
    \).
     
    Last edited: Sep 21, 2008
  2. jcsd
  3. Sep 21, 2008 #2
    Re: Does my "proof" work?

    not diverging may not be equivalent to converging. it may be merely bounded but oscillating. so i would contradict boundedness rather convergence. maybe by showing that the difference between the nth and the n+k is larger than any bound.
     
  4. Sep 21, 2008 #3

    morphism

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    Re: Does my "proof" work?

    But such a sequence is certainly diverging / not converging.


    About the proof presented in the OP: isn't your N_2 redundant? After all, n+1 > n >= N_1. Other than this minor quibble, your proof is fine.
     
  5. Sep 21, 2008 #4
    Re: Does my "proof" work?

    I thought something was wrong with my proof. I was wondering whether N_2 was redundant. I can repeat the definition using N_1 again, or N.

    I was wondering about the contradiction though. Does it really contradict? I proved the opposite based off the assumption, therefore I can assume the original is correct?
     
  6. Sep 21, 2008 #5

    morphism

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    Re: Does my "proof" work?

    You showed that for any convergent sequence (a_n), the quantity |a_{n+1} - a_n| can be made arbitrarily small (by choosing sufficiently large n). This means that a sequence for which this cannot happen is divergent.
     
  7. Sep 21, 2008 #6
    Re: Does my "proof" work?

    ahh, thank you
     
  8. Sep 21, 2008 #7
    Re: Does my "proof" work?

    yeah my first guess is usually wrong. you would think that divergence is always unbounded.
     
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