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Radius of convergence help

  1. Mar 8, 2009 #1
    Hi there - I'm trying to work out the radius of convergence of the series [itex] \sum_{n \geq 1} n^{\sqrt{n}}z^n [/itex] and I'm not really sure where to get going - I've tried using the ratio test and got (not very far) with [itex]lim_{n \to \infty} | \frac{n^{\sqrt{n}}}{(n+1)^{\sqrt{n+1}}}|[/itex], and with the root test, [itex]\left( {lim sup_{n \to \infty} n^{\frac{1}{\sqrt{n}}}}\right) ^{-1} [/itex], neither of which seem to help me =/

    I have a strong feeling the latter converges to 1 but even if I'm right I'm not totally sure how to prove it, and I may well be wrong. What should my next move be?

    Thanks a lot!

    Mathmos6
     
  2. jcsd
  3. Mar 8, 2009 #2

    Dick

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    Don't give up so fast on the root test. What IS the limit n->infinity n^(1/sqrt(n))?
     
  4. Mar 8, 2009 #3
    I'm guessing 1, but I'm not sure how to prove it?
     
  5. Mar 8, 2009 #4

    Dick

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    Take the log to turn it into a quotient. Try to find the limit of the log. Now you can use things like l'Hopital's theorem.
     
  6. Mar 10, 2009 #5
    That's brilliant, thanks! :)
     
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