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Question about a sequence.

  1. Jul 1, 2011 #1
    1. The problem statement, all variables and given/known data
    Determine the convergence or divergence of an = np / en


    3. The attempt at a solution
    Using L'Hopitals Rule, I get (p(nP-1en) - nPen) / e2n which, if I take the limit as n [itex]\rightarrow[/itex][itex]\infty[/itex] I still get [itex]\infty/\infty[/itex] which doesn't help. I can see if a sequence converges to 0, i'm not sure how to show that it does, in general other than taking the limit or using L'Hopital.
     
  2. jcsd
  3. Jul 1, 2011 #2

    Dick

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    For one thing that's not l'Hopital, that's the quotient rule. l'Hopital isn't that. For a second thing I don't think l'Hopital is very helpful anyway. Why don't you take the log of your sequence and try to show the log approaches -infinity. That would show the sequence approaches 0, right?
     
  4. Jul 2, 2011 #3

    Gib Z

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    The form of the sequence is primed for use of the ratio test - if we show the limit of the ratios of consecutive terms is less than 1, then this sequence is eventually smaller than a geometric sequence whose ratio is less than 1, so...
     
  5. Jul 3, 2011 #4
    This is true, I made the mistake of using quotient rule when I meant to apply L'Hopital. I knew it would happen sooner or later :blushing:. Well, it seems so. Again, thanks.
     
  6. Jul 3, 2011 #5
    Gib Z, thank you for your reply. Although I would love to use the ratio test, it so happens that I just haven't gone that far into the chapter to know how to use the ratio test. Coming attractions I suppose :tongue2: thanks for the insight!
     
  7. Jul 3, 2011 #6

    Dick

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    Did you try showing log(n^p/e^n) approaches -infinity? Use that log(n)/n approaches 0. You can prove that with l'Hopital.
     
  8. Jul 3, 2011 #7
    an = pn/en
    ln(an) = nln(p)/nln(e) [lne = 1 so i'm going to just write n here]
    ln(an) = nln(p)/n
    ln(an) = ln(p)
    lim[n -> [itex]\infty[/itex] ]ln(an) = lim[n -> [itex]\infty[/itex] ](ln(p))

    my problem, something i'm not seeing or something I have messed up, is that my term on the right * ln(p) * does nothing when I take the limit [ n -> [itex]\infty[/itex] ]
     
  9. Jul 4, 2011 #8

    Dick

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    log(a/b) isn't log(a)/log(b). It's log(a)-log(b).
     
  10. Jul 5, 2011 #9
    okay, so I have
    {an} = np/en, p > 0
    ln{an} = pln(n) - n p > 0
    lim[n -> [itex]\infty[/itex]]{an} = lim[n -> [itex]\infty[/itex]]pln(n) - lim[n -> [itex]\infty[/itex]] n p > 0
    lim[n -> [itex]\infty[/itex]]{an} = lim[n -> [itex]\infty[/itex]]pln(n) - [itex]\infty[/itex] p > 0
    i'm sorry, i'm really not seeing where this is going. I was thinking of working backwards once I showed {an} converged to 0.
     
  11. Jul 5, 2011 #10

    Dick

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    Write p*log(n)-n as n*(p*log(n)/n-1). As n->infinity how does log(n)/n behave?
     
  12. Jul 5, 2011 #11
    I believe it converges to zero.
     
  13. Jul 5, 2011 #12

    Dick

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    Use l'Hopital to show it does.
     
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