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Limit of Sequence

  1. Jun 12, 2012 #1
    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?

    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?
     
    Last edited: Jun 12, 2012
  2. jcsd
  3. Jun 12, 2012 #2

    SammyS

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    For (1):

    Notice that [itex]\displaystyle \frac{e^n}{\pi^{n/2}}=\left(\frac{e}{\sqrt{\pi}}\right)^{n}[/itex]
     
  4. Jun 12, 2012 #3
    yeah lol I realized after I posted, hence the edit :D
    any ideas about the second one?
    not sure what I've done wrong
     
  5. Jun 12, 2012 #4

    SammyS

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

    [itex]\displaystyle n\,\ln\left(\frac{2}{n}\right)=n\left(\ln(2)-\ln(n)\right)[/itex]

    What's that limit as n → ∞ ?
     
  6. Jun 12, 2012 #5
    Wouldnt that just be infinity :S
     
  7. Jun 13, 2012 #6

    SammyS

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    More like -∞ .

    But that's the exponent on e ... so that gives ____ ?
     
  8. Jun 13, 2012 #7
    ohhh I see
    so it equals 0, thanks!
     
  9. Jun 13, 2012 #8
    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?
     
  10. Jun 13, 2012 #9

    SammyS

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    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 [itex]\displaystyle \left(\lim_{n\to\infty}\frac{2}{n}\right)^{ \displaystyle \left( \lim_{n\to\infty} \,n\right)}\ .[/itex] [/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.
     
    Last edited: Jun 13, 2012
  11. Jun 13, 2012 #10
    Hi SammyS! :smile:

    I'm quite sure that [itex]0^{\infty}[/itex] is not indeterminate, and is equal to 0.
     
  12. Jun 13, 2012 #11

    SammyS

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    Yes, you are correct!

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

    GOOD CATCH !
     
  13. Jun 13, 2012 #12
    so what I have done is okay?
     
  14. Jun 13, 2012 #13

    SammyS

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