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Math Help - Covergence of series

  1. #1
    Senior Member Twig's Avatar
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    Covergence of series

    Hi


    I want to determine if this series is convergent or divergent.
    Can anyone give me a hint on which theorem to use here?

     \displaystyle\sum_{k=1}^{\infty} \; ln(1+\frac{1}{k})

    thx
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  2. #2
    Senior Member Twig's Avatar
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    Can I use  ln(1+\frac{1}{k}) \geq ln(\frac{1}{k})

    And \displaystyle\sum_{k=1}^{\infty} \; ln(\frac{1}{k}) is divergent , so my original series is also diverging?
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  3. #3
    Senior Member Spec's Avatar
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    \ln\left(\frac{1}{k}\right)\leq \ln\left(1+\frac{1}{k}\right)

    Study \ln\left(\frac{1}{k}\right)=\ln(1)-\ln(k) and see that it diverges.

    EDIT: Uhm, you got it.
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  4. #4
    Moo
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    Hello,
    Quote Originally Posted by Twig View Post
    Hi


    I want to determine if this series is convergent or divergent.
    Can anyone give me a hint on which theorem to use here?

     \displaystyle\sum_{k=1}^{\infty} \; ln(1+\frac{1}{k})

    thx
    1+\frac 1k=\frac{k+1}{k}

    So \ln\left(1+\frac 1k\right)=\ln(k+1)-\ln(k)

    So this is a telescoping series and finally, \sum_{k=1}^\infty \ln\left(1+\frac 1k\right)=\lim_{n\to\infty} \sum_{k=1}^n \ln\left(1+\frac 1k\right)=\lim_{n\to\infty} \ln(n+1)-\ln(1)=\infty
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  5. #5
    MHF Contributor chisigma's Avatar
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    In general the series...

    \sum_{n=1}^{\infty} \ln (1+a_{n})

    ... converges if and only if converges the series...

    \sum_{n=1}^{\infty} a_{n}

    Kind regards

    \chi \sigma
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  6. #6
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    Hello chisigma,

    I thought  \frac {1}{k} converges?
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  7. #7
    Senior Member Twig's Avatar
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    Quote Originally Posted by calc101 View Post
    Hello chisigma,

    I thought  \frac {1}{k} converges?

    Itīs diverging. The n:th term approaches zero as n approaches infinity, but

     \displaystyle\sum_{k=1}^{\infty} \frac{1}{k} is diverging.

    You know about the integral test?

     \displaystyle\lim_{R\to \infty} \int_{1}^{R} \frac{1}{x} \, dx = ln(R)-ln(1)  = \infty


    Note:  \displaystyle\sum_{k=1}^{\infty} \frac{1}{k^{1.000001}} is convergent.
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  8. #8
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    No, I meant the sequence \frac {1}{k} converges to 0?

    Not the function \frac {1}{x}
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  9. #9
    Senior Member Twig's Avatar
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    Yes, the sequence  (a_{n})_{n=1}^{\infty} = \frac{1}{n} converges to 0.
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