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Convergence/Divergence of (-1)^n(ln(n)/root(n))

  1. Feb 4, 2012 #1
    1. The problem statement, all variables and given/known data

    Σ (-1)^n(ln(n)/root(n))

    2. Relevant equations

    None

    3. The attempt at a solution

    Tried all the tests, and the only one that makes sense to use is the Alternating Serioes Test. I know the lmit of ln (n)/ root (n) approaches 0. But not sure how to prove that ln (n) / root n is greater than ln n+1 / root (n+1)

    I know the series converges, but not sure of the process in finding so.
     
  2. jcsd
  3. Feb 4, 2012 #2

    SammyS

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    Hello Somnus. Welcome to PF !

    I'm pretty sure that the series does not converge.
     
  4. Feb 4, 2012 #3
    According to the answers in my book, the series converges.
     
  5. Feb 4, 2012 #4

    Dick

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    To apply the alternating series test you could show that for x large enough that the derivative of ln(x)/sqrt(x) is negative, so it's a decreasing function. It does converge.
     
    Last edited: Feb 4, 2012
  6. Feb 4, 2012 #5

    SammyS

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    OK ! I stand corrected.

    I'll take Dick's word over that of WolframAlpha. I had based my earlier comment on a result from WolframAlpha.

    The convergence must be awfully "slow".
     
  7. Feb 5, 2012 #6

    Dick

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    It's not absolutely convergent if that's what you mean by slow. WA is easy to confuse if you don't think about the question you send it.
     
  8. Feb 5, 2012 #7

    SammyS

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  9. Feb 5, 2012 #8

    Deveno

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    Wolfram|Alpha appears to be just wrong on this one (and it's not a matter of "bad input").

    provided one has shown that lim n→∞ an = 0 (and L'Hopital's Rule works well here), than Liebnitz's test applies, and to show monoticity, one can examine the derivative of:

    a(x) = ln(x)/√x.

    it's not hard to show that there is an N such that an is montonically decreasing for all n > N (N is even fairly small).

    one caveat: the starting integer is not explicitly given, although given the nature of the series, 2 seems a reasonable assumption.
     
  10. Feb 5, 2012 #9

    Dick

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    I have seen WA announce that a divergent series is convergent, giving a finite sum by reordering the terms. Perhaps here it is reordering the terms to turn a convergent series into a divergent one. I think it plays a little loosely with conditionally convergent series.
     
  11. Feb 5, 2012 #10

    Deveno

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    i think it's just a bad algorithm somewhere (it's probably comparing it to some other divergent series, and mis-matching the terms. for conditionally convergent series, the way you sum the series matters very much). the fact that the first non-zero term is a2 may have something to do with this. i sent WA an e-mail, they may investigate, they may not.
     
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