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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.

  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.

  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

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    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]

    Last edited: May 12, 2010
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