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Math Help - prove that series

  1. #1
    Super Member dhiab's Avatar
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    prove that series

    prove that :
    \sum_{n=1}^{\infty }(\frac{1}{n^{2}})=\frac{\pi ^{2}}{6}
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  2. #2
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    Quote Originally Posted by dhiab View Post
    prove that :
    \sum_{n=1}^{\infty }(\frac{1}{n^{2}})=\frac{\pi ^{2}}{6}

    Consider  \sin(x) = x \prod_{n=1}^{\infty}  \left ( 1  - \frac{x^2}{(n\pi)^2} \right )

    Then compare the coefficient of  x^2  with its Taylor's series .
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  3. #3
    MHF Contributor chisigma's Avatar
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    Quote Originally Posted by simplependulum View Post
    Consider  \sin(x) = x \prod_{n=1}^{\infty} \left ( 1 - \frac{x^2}{(n\pi)^2} \right )

    Then compare the coefficient of  x^2 with its Taylor's series .
    ...  x^2 or  x^3 ? ...

    Kind regards

    \chi \sigma
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  4. #4
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    Oh , yes

    If i say comparing with the series , it should be  x^3 , but not  x^2 .

    Or say compare the coefficient of  x^2 after dividing the factor  x
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  5. #5
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    Quote Originally Posted by dhiab View Post
    prove that :
    \sum_{n=1}^{\infty }(\frac{1}{n^{2}})=\frac{\pi ^{2}}{6}

    Take a peek at this very nice paper: http://www.math.titech.ac.jp/~inoue/...lder/zeta2.pdf
    You have there no less than 14 different proofs of the result \zeta(2)=\sum\limits_{n+1}^{\infty}\frac{1}{n^2}=\  frac{\pi^2}{6}, from rather basic ones to those that require more knowledge....enjoy!

    Tonio
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  6. #6
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  7. #7
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    I believe the OP mostly posts challenge problems but places them in the regular forums. This thread has been answered many times over and I am going to PM the OP about using the Challenge Problems forum to save everyone time.
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