# Limit of Sequence

1. Jun 12, 2012

### dan38

1. The problem statement, all variables and given/known data
1) e^n / pi^(n/2)
2) (2/n)^n

2. Relevant equations

3. The attempt at a solution

1) Take out 1/pi^0.5 as a factor

Now have limit of (e/pi)^n

Since this ratio is less than 1, it will converge?

2) e ^ limit ( n* ln (2/n))

e ^ limit ( ln(2/n) / ( 1 / n))

Apply l'hopital's rule, end up with e ^ n
Hence it will diverge?

Last edited: Jun 12, 2012
2. Jun 12, 2012

### SammyS

Staff Emeritus
For (1):

Notice that $\displaystyle \frac{e^n}{\pi^{n/2}}=\left(\frac{e}{\sqrt{\pi}}\right)^{n}$

3. Jun 12, 2012

### dan38

yeah lol I realized after I posted, hence the edit :D
any ideas about the second one?
not sure what I've done wrong

4. Jun 12, 2012

### SammyS

Staff Emeritus
$\displaystyle \lim_{n\to\infty} \frac{\ln(2/n}{1/n}$ is of the form $\displaystyle \frac{-\infty}{0}\,,$ so L'Hôpital's rule can't be applied.

$\displaystyle n\,\ln\left(\frac{2}{n}\right)=n\left(\ln(2)-\ln(n)\right)$

What's that limit as n → ∞ ?

5. Jun 12, 2012

### dan38

Wouldnt that just be infinity :S

6. Jun 13, 2012

### SammyS

Staff Emeritus
More like -∞ .

But that's the exponent on e ... so that gives ____ ?

7. Jun 13, 2012

### dan38

ohhh I see
so it equals 0, thanks!

8. Jun 13, 2012

### dan38

just thought of something else; would this way work?

lim (2/n)^n
n---> Infinity

(lim 2/n)^n

lim 2/n = 0

(lim 2/n)^n = 0 as well?

9. Jun 13, 2012

### SammyS

Staff Emeritus
No, you can't do that.

You have taken the n which is an exponent out of the limit.

[STRIKE]You can't do it even if you look at it as $\displaystyle \left(\lim_{n\to\infty}\frac{2}{n}\right)^{ \displaystyle \left( \lim_{n\to\infty} \,n\right)}\ .$ [/STRIKE]

[STRIKE]That's of the indeterminate form 0.[/STRIKE]

See the following post. Infinitum is correct, 0 is not of indeterminate form.

Last edited: Jun 13, 2012
10. Jun 13, 2012

### Infinitum

Hi SammyS!

I'm quite sure that $0^{\infty}$ is not indeterminate, and is equal to 0.

11. Jun 13, 2012

### SammyS

Staff Emeritus
Yes, you are correct!

I had just come back to Edit my post, when I saw yours.

GOOD CATCH !

12. Jun 13, 2012

### dan38

so what I have done is okay?

13. Jun 13, 2012

### SammyS

Staff Emeritus
Well, you should also have a limit on your exponent.