1. Jul 1, 2011

### icesalmon

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 $\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.

2. Jul 1, 2011

### 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?

3. Jul 2, 2011

### 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...

4. Jul 3, 2011

### icesalmon

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.

5. Jul 3, 2011

### icesalmon

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. Jul 3, 2011

### Dick

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. Jul 3, 2011

### icesalmon

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 -> $\infty$ ]ln(an) = 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 -> $\infty$ ]

8. Jul 4, 2011

### Dick

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

9. Jul 5, 2011

### icesalmon

okay, so I have
{an} = np/en, p > 0
ln{an} = pln(n) - n p > 0
lim[n -> $\infty$]{an} = lim[n -> $\infty$]pln(n) - lim[n -> $\infty$] n p > 0
lim[n -> $\infty$]{an} = 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 {an} converged to 0.

10. Jul 5, 2011

### Dick

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

11. Jul 5, 2011

### icesalmon

I believe it converges to zero.

12. Jul 5, 2011

### Dick

Use l'Hopital to show it does.