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Sequence limit proof

  1. Oct 6, 2011 #1
  2. jcsd
  3. Oct 6, 2011 #2

    SammyS

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    I doubt that he wants an ε - N proof.

    Probably wants more than a rough demonstrative "proof". I would.

    Do a comparison test. The hint is right there !
     
  4. Oct 6, 2011 #3
    Oh, I know. I just thought that was too easy. I'll work it out shortly.
     
  5. Oct 6, 2011 #4
    xn = (sn)

    sn+1/sn = xn+1/xn = x*xn/xn = x

    x ≥ L > 1

    Eh. I think that "proves" it. Assuming x > 1, then the sequence ratio is bigger than 1. And, of course, the limit is given which is infinity which is what we were asked to show for a sequence that satisfies that ratio.
     
    Last edited: Oct 6, 2011
  6. Oct 6, 2011 #5

    SammyS

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    I It doesn't say xn = (sn) at all.

    sn+1 ≥ L sn ≥ L2 sn-1 ≥ L3 sn-2 ≥ ...

    What can you say about [itex]\displaystyle \lim_{n\to\infty}L^n\,?[/itex]

    This leads nicely to an ε - N proof. Maybe do one.
     
  7. Oct 6, 2011 #6
    That limit should go to infinity, right?

    The book does the proof for when L < 1. I'm looking over that proof now.
     
    Last edited: Oct 6, 2011
  8. Oct 6, 2011 #7

    SammyS

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    No, it doesn't go to zero. L > 1 !
     
  9. Oct 6, 2011 #8
    Yeah, I didn't mean to put zero.
     
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