Does the Sequence an = np / en Converge or Diverge?

[icesalmon]

Homework Statement

Determine the convergence or divergence of a_n = n^p / eⁿ

The Attempt at a Solution

Using L'Hopitals Rule, I get (p(n^P-1eⁿ) - n^Peⁿ) / e²ⁿ which, if I take the limit as n \rightarrow\infty I still get \infty/\infty 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.

 

[Dick]

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?

 

[Gib Z]

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...

 

[icesalmon]

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 . Well, it seems so. Again, thanks.

 

[icesalmon]

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 thanks for the insight!

 

[Dick]

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 . 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.

 

[icesalmon]

a_n = pⁿ/eⁿ
ln(a_n) = nln(p)/nln(e) [lne = 1 so I'm going to just write n here]
ln(a_n) = nln(p)/n
ln(a_n) = ln(p)
lim[n -> \infty ]ln(a_n) = lim[n -> \infty ](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 -> \infty ]

 

[Dick]

a_n = pⁿ/eⁿ
ln(a_n) = nln(p)/nln(e) [lne = 1 so I'm going to just write n here]
ln(a_n) = nln(p)/n
ln(a_n) = ln(p)
lim[n -> \infty ]ln(a_n) = lim[n -> \infty ](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 -> \infty ]

log(a/b) isn't log(a)/log(b). It's log(a)-log(b).

 

[icesalmon]

okay, so I have
{a_n} = n^p/eⁿ, p > 0
ln{a_n} = pln(n) - n p > 0
lim[n -> \infty]{a_n} = lim[n -> \infty]pln(n) - lim[n -> \infty] n p > 0
lim[n -> \infty]{a_n} = lim[n -> \infty]pln(n) - \infty p > 0
i'm sorry, I'm really not seeing where this is going. I was thinking of working backwards once I showed {a_n} converged to 0.

 

[Dick]

okay, so I have
{a_n} = n^p/eⁿ, p > 0
ln{a_n} = pln(n) - n p > 0
lim[n -> \infty]{a_n} = lim[n -> \infty]pln(n) - lim[n -> \infty] n p > 0
lim[n -> \infty]{a_n} = lim[n -> \infty]pln(n) - \infty p > 0
i'm sorry, I'm really not seeing where this is going. I was thinking of working backwards once I showed {a_n} converged to 0.

Write p*log(n)-n as n*(p*log(n)/n-1). As n->infinity how does log(n)/n behave?

 

[icesalmon]

I believe it converges to zero.

 

[Dick]

I believe it converges to zero.

Use l'Hopital to show it does.