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Question about a sequence.

  • Thread starter icesalmon
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  • #1
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Homework Statement


Determine the convergence or divergence of an = np / en


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.
 

Answers and Replies

  • #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?
 
  • #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...
 
  • #4
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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?
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.
 
  • #5
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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...
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!
 
  • #6
Dick
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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.
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.
 
  • #7
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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] ]
 
  • #8
Dick
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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] ]
log(a/b) isn't log(a)/log(b). It's log(a)-log(b).
 
  • #9
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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.
 
  • #10
Dick
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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.
Write p*log(n)-n as n*(p*log(n)/n-1). As n->infinity how does log(n)/n behave?
 
  • #11
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I believe it converges to zero.
 
  • #12
Dick
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I believe it converges to zero.
Use l'Hopital to show it does.
 

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