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Good dayQuestion: Determine whether the series is convergent or

  1. Feb 2, 2012 #1
    Good day..

    Question: Determine whether the series is convergent or divergent:

    Series starts at n=1 and goes to infinity.. Of 2/(n*4throot(2n+2))

    What I mean is.. 2/(n*(2n+2)^(1/4))

    Can someone tell me which test to try? I cant get it in the form of a p-series.. so I think maybe the Integral test would be worth a shot?
     
  2. jcsd
  3. Feb 2, 2012 #2

    jbunniii

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    Re: Convergence/Divergence

    Can you compare it to a p-series?
     
  4. Feb 2, 2012 #3
    Re: Convergence/Divergence

    Well it sort of looks like it but I can't get it in a form to be confident with an answer
     
  5. Feb 2, 2012 #4

    jbunniii

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    Re: Convergence/Divergence

    Hint:

    [tex]\sum_{n=1}^{\infty} \frac{1}{n*(n)^{1/4}}[/tex]

    is a p-series. Does it converge? Can you compare your series to it?
     
  6. Feb 2, 2012 #5
    Re: Convergence/Divergence

    In your example you would get 1/n^(5/4) where p = 5/4 >1 so that would converge.. correct?

    I realize mine could be similar... Ʃ[ 2/(n*(2n+2)^(1/4))] But I can't combine the n's on the bottom because the +2 is messing with me.
     
  7. Feb 2, 2012 #6

    jbunniii

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    Re: Convergence/Divergence

    Correct.

    How does [itex]1/(2n+2)^{1/4}[/itex] compare with [itex]1/n^{1/4}[/itex]? Which one is bigger?
     
  8. Feb 3, 2012 #7
    Re: Convergence/Divergence

    I would like to this 1/(2n+2)^(1/4) is bigger.
     
  9. Feb 3, 2012 #8
    Re: Convergence/Divergence

    Well actually 1/(2n+2)^(1/4) would go to zero faster so I suppose it's smaller?
     
  10. Feb 3, 2012 #9

    jbunniii

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    Re: Convergence/Divergence

    Right. So can you use this fact to apply the comparison test, and conclude that the series converges?
     
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