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Math Help - closed form of nth partial sum

  1. #1
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    closed form of nth partial sum

    find a closed form for the nth partial sum of the series and determine whether the series converges. If so, find its sum.

    ln(1 - (1/4)) + ln(1 - (1/9)) + ln(1 - (1/16)) +...+ ln(1 - (1/(k+1)^2)+...

    using log properties i got the first few partial sums as S1 = 3/4, S2 = 2/3, S3 = 5/8, S4 = 3/5 but i can't seem to find a pattern. also in the answers in the back of the book, even though this question asked for a closed form, the back of the book gave: sum from k=2 to (n+1) of [ln((k-1)/(k)) - ln(k/(k+1))]. how did the book get that answer?
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by oblixps View Post
    find a closed form for the nth partial sum of the series and determine whether the series converges. If so, find its sum.

    ln(1 - (1/4)) + ln(1 - (1/9)) + ln(1 - (1/16)) +...+ ln(1 - (1/(k+1)^2)+...

    using log properties i got the first few partial sums as S1 = 3/4, S2 = 2/3, S3 = 5/8, S4 = 3/5 but i can't seem to find a pattern. also in the answers in the back of the book, even though this question asked for a closed form, the back of the book gave: sum from k=2 to (n+1) of [ln((k-1)/(k)) - ln(k/(k+1))]. i thought closed form meant without a summation symbol. i am confused.
    \sum_{k=2}^{m}\ln\left(1-\frac{1}{k^2}\right)=\sum_{k=2}^{m}\ln\left(\frac{  k^2-1}{k^2}\right)= \sum_{k=2}^{m}\left\{\ln(k^2-1)-\ln(k^2)\right)=-\sum_{k=2}^{m}\left\{\ln(k^2)-\ln(k^2-1)\right\}
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  3. #3
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    i don't understand how you got that. could you explain how your first step came about?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by oblixps View Post
    i don't understand how you got that. could you explain how your first step came about?
    1-\frac{1}{k^2}=\frac{k^2}{k^2}-\frac{1}{k^2}=\frac{k^2-1}{k^2}
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  5. #5
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    oh wait i understand it now. silly me. thanks!
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  6. #6
    MHF Contributor chisigma's Avatar
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    The exact value of...

    S = \lim_{m \rightarrow \infty} \sum_{k=2}^{m} \ln (1-\frac{1}{k^{2}}) (1)

    ... can be found as follows. Starting from the well known 'infinite product' ...

    \frac{\sin \pi x}{\pi x}= \prod_{k=1}^{\infty} (1-\frac{x^{2}}{k^{2}}) (2)

    ... we obtain first...

    \prod_{k=2}^{\infty} (1-\frac{x^{2}}{k^{2}}) = \frac{\sin \pi x}{\pi x (1-x^{2})} (3)

    ... and then...

     \prod_{k=2}^{\infty} (1-\frac{1}{k^{2}}) = \lim_{x \rightarrow 1} \frac{\sin \pi x}{\pi x (1-x^{2})} = \lim_{x \rightarrow 1} \frac{\cos \pi x}{(1-3 x^{2})} = \frac{1}{2} (4)

    ... so that...

    S = \lim_{m \rightarrow \infty} \sum_{k=2}^{m} \ln (1-\frac{1}{k^{2}})= - \ln 2 (5)

    Kind regards

    \chi \sigma
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