1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Limit of sequence

  1. May 12, 2010 #1
    1. The problem statement, all variables and given/known data
    Prove that
    [tex] \lim_{n \rightarrow \infty} \sqrt[n]{\frac{2^n}{n!!}}=0 [/tex]

    2. Relevant equations


    3. The attempt at a solution
    Seems really tricky ...
     
  2. jcsd
  3. May 12, 2010 #2
    You could try rewriting it in exponential form, getting rid of the root. This will help you see what to do next.



    -F
     
  4. May 12, 2010 #3
    Not sure what you mean with exponential form.
     
  5. May 12, 2010 #4
    If you mean

    [tex] e^{\frac{1}{n} \log (\frac{2^n}{n!!}) } [/tex]

    it's not helping me.
     
  6. May 12, 2010 #5
    How about, [tex]\sqrt[n]{x}[/tex]=[tex]x^{n^{-1}}[/tex]

    This way you might not need logarithms, and you might be able to simplify it enough for it to be clear to you.


    -F
     
  7. May 12, 2010 #6
    It's a little simplified, as I get

    [tex] 2 \sqrt[n]{\frac{1}{n!!}} [/tex]

    .. but what about this term?
     
  8. May 12, 2010 #7

    Char. Limit

    User Avatar
    Gold Member

    Well, what happens to that term as [itex]n -> \infty[/itex]?
     
  9. May 12, 2010 #8
    It hopefully goes to zero but this is not a proof. It is not clear at all because for instance

    [tex] \lim_{n \rightarrow \infty} \sqrt[n]{\frac{1}{n}} = 1 [/tex]

    so...
     
    Last edited: May 12, 2010
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook