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Math Help - Does this series converge or diverge?

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    Does this series converge or diverge?

    Does this series converge or diverge?

    1 / ( (ln(n)) ^ (ln(n)) )

    I'm not really sure how to approach this. It doesn't look like you can solve an indefinite integral, I can't find anything to used for the comparison test, and the ration test doesn't really simplify. What should I do? Thank you!
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    Quote Originally Posted by uberbandgeek6 View Post
    Does this series converge or diverge?

    1 / ( (ln(n)) ^ (ln(n)) )

    I'm not really sure how to approach this. It doesn't look like you can solve an indefinite integral, I can't find anything to used for the comparison test, and the ration test doesn't really simplify. What should I do? Thank you!

    You're forgetting, apparenty, the very nice Cauchy's Condensation Test: if a sequence \{a_n\} is positive and monotonically descending to zero, the series \sum\limits_{n=1}^\infty a_n converges iff the series \sum\limits_{n=1}^\infty 2^na_{2^n} converges (and we can take any prime p instead of 2).
    Apply this to your series and you'll find out it diverges (after, perhaps, you apply the n-th root test to the result of taking 2^na_{2^n})

    Tonio
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    We never really learned that, so I would probably get docked points for not using a technique I already know.
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    Quote Originally Posted by uberbandgeek6 View Post
    We never really learned that, so I would probably get docked points for not using a technique I already know.

    Well, then first prove it. Try to google it or read any decent calculus book.

    Tonio
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    Quote Originally Posted by uberbandgeek6 View Post
    We never really learned that, so I would probably get docked points for not using a technique I already know.
    That is a practice that many people donít understand. But in many schools it is widely practiced. My guess is that your textbook & instructor want you to use the comparison test.

    Here is the trick I learned thanks to Gillmanís text.
    \left(\ln(n)\right)^{\ln(n)}=n^{\ln[ln(n)]}.
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