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Math Help - Series' Convergence?

  1. #1
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    Series' Convergence?

    Hello


    Determine whether the following series is convergent or divergent:

    \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{ln^e(n)}

    This one makes me crazy!
    How to deal with it ?!
    The power e is on the whole ln.
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  2. #2
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    I don't know, ratio seems to be of the chart here, root likewise. Have you tried the integral test?
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  3. #3
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    Ratio will give 1.
    Can not use integral test since \displaystyle \int \dfrac{1}{ln^e(x)} \, dx is an unelemetary integral.
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    Is this true, \dfrac{1}{{\ln ^e (n)}} > \dfrac{1}{{\ln ^3 (n)}}~?

    Use a comparison test.
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    Quote Originally Posted by Liverpool View Post
    Hello


    Determine whether the following series is convergent or divergent:

    \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{ln^e(n)}

    This one makes me crazy!
    How to deal with it ?!
    The power e is on the whole ln.
    It's also fair game to use Cauchy's Condensation test. Namely \displaystyle \frac{2^n}{\log^e\left(2^n\right)}=\frac{2^n}{n^e\  log^e(2)}\to\infty
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  6. #6
    Grand Panjandrum
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    Quote Originally Posted by Liverpool View Post
    Hello


    Determine whether the following series is convergent or divergent:

    \displaystyle \sum_{n=2}^{\infty} \dfrac{1}{ln^e(n)}

    This one makes me crazy!
    How to deal with it ?!
    The power e is on the whole ln.
    I may be being a bit dense here, but what does the notation \ln^e(n) denote? Is this a fractionally iterated logarithm, a fractional derivative (not likely I would have thought), abused notation for (\ln(n))^e, ... ?

    By the way as \ln(n) grows more slowly than any positive power of $$n, so does (\ln(n))^e and in particular for $$n large enough there exists a $$k such that (\ln(n))^e<k\times n, and so the series diverges.


    CB
    Last edited by CaptainBlack; December 29th 2010 at 01:27 AM.
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    Thanks.
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