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Math Help - Is the series covergent or divergent?

  1. #1
    Senior Member Twig's Avatar
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    Is the series covergent or divergent?

    Hi!

    Problem: \sum_{k=1}^{\infty} (\sqrt[k]{k}-1) , convergent or divergent?

    Thx
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  2. #2
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    Use x>\ln(1+x) for x>0 with x=\root k\of k-1. Then \root k\of k-1>\ln(\root k\of k)=\frac{\ln k}k.

    By comparison with \sum\frac1k the series diverges.
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  3. #3
    MHF Contributor
    Jester's Avatar
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    That is clever!
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