Limit of Sequence Homework Solutions

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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?
 
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dan38 said:

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}
 
yeah lol I realized after I posted, hence the edit :D
any ideas about the second one?
not sure what I've done wrong
 
dan38 said:
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 → ∞ ?
 
Wouldnt that just be infinity :S
 
dan38 said:
Wouldnt that just be infinity :S

More like -∞ .

But that's the exponent on e ... so that gives ____ ?
 
ohhh I see
so it equals 0, thanks!
 
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?
 
dan38 said:
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.
 
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  • #10
SammyS said:
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! :smile:

I'm quite sure that 0^{\infty} is not indeterminate, and is equal to 0.
 
  • #11
Infinitum said:
Hi SammyS! :smile:

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 !
 
  • #12
so what I have done is okay?
 
  • #13
dan38 said:
so what I have done is okay?

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