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Math Help - Some problems with series

  1. #16
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Hweengee View Post
    How do I determine the convergence/divergence of the following:

     <br />
\sum _{n=1}^{\infty }\left({\sqrt[n] 3}-1 \right)<br />

    I can't use limit comparison test with {\sqrt[n] 3} because it gives me 0. Root and ratio tests don't seem appropriate and I'm struggling to find a divergent series which is smaller than this to compare it with. Would it be possible to compare it with \sum _{n=1}^{\infty }\left(\ln\left(n \right)-1 \right)?

    As for

    \sum _{n=3}^{\infty }\frac{1}{\left( lnln\left(n \right) \right)^{\ln \left( n \right) }}

    I'm more or less completely stuck.

    Thanks for the help in advance.
    I see that your second series was never actually solved. Well I would suggest Cauchy's Condensation test. Since the sequence in the series is positive and decreasing it only converges iff \sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2^n))^{\ln(2  ^n)}}=\sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2)n)^{  \ln(2)n}}\sim\sum_{n=3}^{\infty}\frac{2^n}{\ln(n)^  n}. Now it is pretty clear this converges.
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  2. #17
    Junior Member toraj58's Avatar
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    Question

    would you please explain furthur that why it converge. as a glance i see that it is divergant.
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  3. #18
    Junior Member toraj58's Avatar
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    Thumbs up

    sorry with te RATIO TEST i found it convergant
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  4. #19
    MHF Contributor

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    Quote Originally Posted by Mathstud28 View Post
    I see that your second series was never actually solved. Well I would suggest Cauchy's Condensation test. Since the sequence in the series is positive and decreasing it only converges iff \sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2^n))^{\ln(2  ^n)}}=\sum_{n=3}^{\infty}\frac{2^n}{\ln(\ln(2)n)^{  \ln(2)n}}\sim\sum_{n=3}^{\infty}\frac{2^n}{\ln(n)^  n}. Now it is pretty clear this converges.
    It is possible to compare with a simple series: we have (\ln\ln n)^{\ln n}=e^{\ln n\ln\ln\ln n}=n^{\ln\ln\ln n}\geq n^2 as soon as n\geq e^{e^{e^2}}\simeq 5.8\cdot 10^{702} (!!), hence \frac{1}{(\ln\ln n)^{\ln n}}\leq \frac{1}{n^2} for these pretty large n. Hence it converges.
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