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Math Help - Series correction

  1. #1
    Junior Member simplysparklers's Avatar
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    Series correction

    Hiya guys!
    Could someone correct this for me please?Not sure if I went about it right at all!

    Determine the sum of the series \sum_{n=1}^{\infty}\frac{1}{n(n+2)}

    What I did so far was:
    \sum_{n=1}^{\infty}\frac{1}{n(n+2)}=\frac{1}{1\cdo  t{3}}+\frac{1}{2\cdot{4}}+\frac{1}{3\cdot{5}}+...

    The partial sum:
    S_{n}=\frac{1}{1\cdot{3}}+\frac{1}{2\cdot{4}}+\fra  c{1}{3\cdot{5}}+...+\frac{1}{(n-2)n}+\frac{1}{n(n+2)}

    Rewritten as:

    (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{(n-2)}-\frac{1}{n})+(\frac{1}{n}-{1}{(n+2)})
    (I don't really know why I did this to be honest, I was mostly copying a similar sum from the book, so more than likely gone wrong with that line)

    S_{n}=1+\frac{1}{2}+\frac{1}{(n-2)}+\frac{1}{n}

    S_{n}=\frac{(2)(n-2)(n)+(n-2)(n)+(2)(n)+(2)(n-2)}{(2)(n-2)(n)}
    S_{n}=\frac{3n^{2}-2n-4}{2n^{2}-4n}

    So...did I go completely wrong??Or where do I go from here to prove it's convergent and what S_{n}\rightarrow{??} is??

    Thanks for your help guys!!
    Jo
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  2. #2
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by simplysparklers View Post
    Rewritten as:

    (1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{(n-2)}-\frac{1}{n})+(\frac{1}{n}-{1}{(n+2)})
    (I don't really know why I did this to be honest, I was mostly copying a similar sum from the book, so more than likely gone wrong with that line)
    You knew it all along

    Always try to find the partial fraction decomposition in these cases:

    Here \frac1{n(n+2)} = \frac12 \left(\frac1{n} - \frac1{n+2}\right). So your sum should have read:

    \frac12\left((1-\frac{1}{3})+(\frac{1}{2}-\frac{1}{4})+(\frac{1}{3}-\frac{1}{5})+(\frac{1}{4}-\frac{1}{6})+...+(\frac{1}{(n-2)}-\frac{1}{n})+(\frac{1}{n}-\frac{1}{(n+2)})\right)
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  3. #3
    Junior Member simplysparklers's Avatar
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    Thanks Isomorphism!
    Just wondering though, did I still pick the right terms left over from the series?Does it come out then as:
    S_{n}=\frac{1}{2}(1+\frac{1}{2}+\frac{1}{(n-2)}+\frac{1}{n})?

    Which goes to:
    \frac{3n^{2}-2n-4}{4n^{2}-8n}?

    Thanks,
    Jo
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  4. #4
    Lord of certain Rings
    Isomorphism's Avatar
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    Quote Originally Posted by simplysparklers View Post
    Hiya guys!
    Could someone correct this for me please?Not sure if I went about it right at all!

    Determine the sum of the series \sum_{n=1}^{\infty}\frac{1}{n(n+2)}

    Most likely it is wrong since you did not consider the n+1 term...

    But anyway I know a way to do it...

    \sum_{n=1}^{\infty}\frac{1}{n(n+2)} = \frac12 \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+2} = \frac12 \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+1} + \frac{1}{n+1} - \frac{1}{n+2}

    \frac12 \sum_{n=1}^{\infty}\frac{1}{n} - \frac{1}{n+1} +\frac12 \sum_{n=1}^{\infty} \frac{1}{n+1} - \frac{1}{n+2} = \frac12 \lim_{n \to \infty} 1 - \frac1{n+1} + \frac12 - \frac1{n+2} = \frac34


    EDIT: Your answer is mostly correct since you get the same limit.
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  5. #5
    Junior Member simplysparklers's Avatar
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    Hugs for Isomorphism!!! Thank you!!
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  6. #6
    Math Engineering Student
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    From MacLaurin series:

    \sum\limits_{n\,=\,1}^{\infty }{\frac{1}{n(n+2)}}=\int_{0}^{1}{\left\{ x\sum\limits_{n\,=\,1}^{\infty }{\frac{x^{n}}{n}} \right\}\,dx}=-\int_{0}^{1}{x\ln (1-x)\,dx},

    and we can kill this with standard techniques.
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