Limit of Sequence Homework Solutions

[dan38]

Homework Statement

1) e^n / pi^(n/2)
2) (2/n)^n

Homework Equations

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?

Ooops, dw about this one, realized my mistake ^^

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?

 

[SammyS]

Homework Statement

1) e^n / pi^(n/2)
2) (2/n)^n

Homework Equations

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?

Ooops, dw about this one, realized my mistake ^^

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?

For (1):

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

 

[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

 

[SammyS]

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?

$\displaystyle \lim_{n\to\infty} \frac{\ln(2/n}{1/n}$ 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 → ∞ ?

 

[dan38]

Wouldnt that just be infinity :S

 

[SammyS]

Wouldnt that just be infinity :S

More like -∞ .

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

 

[dan38]

ohhh I see
so it equals 0, thanks!

 

[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?

 

[SammyS]

just thought of something else; would this way work?

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

(lim 2/n)^nlim 2/n = 0

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

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]

Added in Edit:

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

 

[Infinitum]

No, you can't do that.

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

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

That's of the indeterminate form 0^∞.

Hi SammyS!

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

 

[SammyS]

Hi SammyS!

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

Yes, you are correct!

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

GOOD CATCH !

 

[dan38]

so what I have done is okay?

 

[SammyS]

so what I have done is okay?

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