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Math Help - covergense root question mmn11 2c

  1. #1
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    covergense root question mmn11 2c

    \sum_{k=1}^{infinity}ln\frac{\sqrt{k}+1}{\sqrt{k}}
    is it converges absolutly ,or with condition(dont know the english term),
    or is it diverges?
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  2. #2
    Ted
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    \dispalystyle ln \left( \frac{\sqrt{k}+1}{\sqrt{k}} \right) = ln(\sqrt{k}+1) - ln(\sqrt{k})

    so your series can be rewritten as \displaystyle \sum_{k=1}^{\infty} \left( ln(\sqrt{k}+1) - ln(\sqrt{k}) \right)

    Telescoping Series.
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  3. #3
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    by what test its absolut value converging?
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  4. #4
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by Ted View Post
    \dispalystyle ln \left( \frac{\sqrt{k}+1}{\sqrt{k}} \right) = ln(\sqrt{k}+1) - ln(\sqrt{k})

    so your series can be rewritten as \displaystyle \sum_{k=1}^{\infty} \left( ln(\sqrt{k}+1) - ln(\sqrt{k}) \right)

    Telescoping Series.
    \displaystyle \sum_{k=1}^{\infty} \ln \frac{1+\sqrt{k}}{\sqrt{k}}= \ln 2 - \ln 1 + \ln (1+\sqrt{2}}) - \ln {\sqrt{2}} + \ln (1+\sqrt{3}}) - \ln {\sqrt{3}})+ ... (1)

    Is (1) 'telescopic'?...

    Kind regards

    \chi \sigma
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  5. #5
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by transgalactic View Post
    \sum_{k=1}^{infinity}ln\frac{\sqrt{k}+1}{\sqrt{k}}
    is it converges absolutly ,or with condition(dont know the english term),
    or is it diverges?
    For k 'large enough' is...

    \displaystyle \ln \frac{1+\sqrt{k}}{\sqrt{k}} = \ln (1+\frac{1}{\sqrt{k}}) > \frac{1}{k}

    ... so that...

    Kind regards

    \chi \sigma
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  6. #6
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    why ln(1+x)>x
    ??

    1/k diverges
    so it cannot be absolutly convergent
    only conditionaly
    for this series to convers it 1. need to make a1>a2>...>ak i dont see it here

    2.
    and lim a_k =0

    the limit is zero

    how to prove 1?
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  7. #7
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    Quote Originally Posted by chisigma View Post
    \displaystyle \sum_{k=1}^{\infty} \ln \frac{1+\sqrt{k}}{\sqrt{k}}= \ln 2 - \ln 1 + \ln (1+\sqrt{2}}) - \ln {\sqrt{2}} + \ln (1+\sqrt{3}}) - \ln {\sqrt{3}})+ ... (1)

    Is (1) 'telescopic'?...

    Kind regards

    \chi \sigma
    cant see how the numbers cancel out each other
    cant see two similar members
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