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Thread: how to you show the series sigma n^(3/n) converge or diverge?

  1. July 7th 2010, 09:01 PM #1
    de20
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    how to you show the series sigma n^(3/n) converge or diverge?

    How to you show the series sigma n^(3/n) converge or diverge?

    Should i use the com parsing test?
    Last edited by mr fantastic; July 7th 2010 at 09:49 PM. Reason: Copied title into main body of post.
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  2. July 7th 2010, 09:09 PM #2
    pickslides
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    Read the section on "Ratio Test"

    Pauls Online Notes : Calculus II - Ratio Test
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  3. July 7th 2010, 09:15 PM #3
    de20
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    ratio test is inconclusive with this case since lim n->inf n^[(3/n+1)-3/n] =1
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  4. July 7th 2010, 09:46 PM #4
    roninpro
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    The series fails the n-th term test.

    \lim_{n\to \infty} n^{3/n}=1
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  5. July 7th 2010, 09:47 PM #5
    mr fantastic
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    Quote Originally Posted by de20 View Post
    should i use the com parsing test?
    Since \displaystyle {\lim_{n \to +\infty} n^{3/n} = 1} the series is clearly divergent by the limit test.
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