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Math Help - Express in closed form

  1. #1
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    Express in closed form

    Im trying to find a closed form for the nth partial sum of the series ln(1/2)+ln(2/3)+ln(3/4)+...+ln(k/(k+1))+....
    My book only has a paragraph on this and i cant understand what they're doing, can someone please try explaing this to me a little better?
    Thanks
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  2. #2
    Super Member PaulRS's Avatar
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    Ok you want to compute: \sum_{k=1}^n{\ln\left(\frac{k}{k+1}\right)}

    Note that \ln\left(\frac{k}{k+1}\right)=\ln(k)-\ln(k+1)

    So: \sum_{k=1}^n{\left(\ln(k)-\ln(k+1)\right)} =[\ln(1)-\ln(2)]+[\ln(2)-\ln(3)]...+[\ln(n-1)-\ln(n)]+[\ln(n)-\ln(n+1)]

    It's a telescoping sum , thus shaking off we have: \sum_{k=1}^n{\left(\ln(k)-\ln(k+1)\right)}=-\ln(n+1)
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  3. #3
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    That was fast! Thanks for the help
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  4. #4
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    Hello, cowboys111!

    A slightly different approach . . .


    Find a closed form for the n^{th} partial sum of the series:
    . . \ln\left(\frac{1}{2}\right)+\ln\left(\frac{2}{3}\r  ight)+\ln\left(\frac{3}{4}\right) + \ln\left(\frac{4}{5}\right) + \hdots

    We have: . S \;=\;\ln\left(\frac{1}{2}\right) + \ln\left(\frac{2}{3}\right) + \ln\left(\frac{3}{4}\right) + \hdots + \ln\left(\frac{n}{n+1}\right)

    . . S \;=\;\ln\left(\frac{1}{2}\cdot\frac{2}{3}\cdot\fra  c{3}{4}\cdots\frac{n}{n+1}\right) \;=\;\ln\left(\frac{1}{n+1}\right) \;=\; \ln(n+1)^{-1}


    Therefore: . S \;=\;-\ln(n+1)

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