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Math Help - Infinite series. (14)

  1. #1
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    Infinite series. (14)

    Hello
    Find the sum of the following series :
    1- \sum_{n=0}^{\infty} \frac{(-1)^n {\pi}^n}{9^n (2n)!}
    2- \sum_{n=2} ln(1-\frac{1}{n^2})

    hmmm I do not have any ideas
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  2. #2
    Super Member General's Avatar
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    Quote Originally Posted by TWiX View Post
    Hello
    Find the sum of the following series :
    1- \sum_{n=0}^{\infty} \frac{(-1)^n {\pi}^n}{9^n (2n)!}
    2- \sum_{n=2} ln(1-\frac{1}{n^2})

    hmmm I do not have any ideas

    1-
    \color{blue}{\pi}^n=(\sqrt{\pi})^{2n}
    \color{blue}9^n=3^{2n}
    Now, think about a well-known maclaurin series.

    2-
    Hint:

    \color{blue}ln(1-\frac{1}{n^2})=ln(\frac{n^2-1}{n^2})=ln(\frac{(n-1)(n+1)}{(n)(n)})=ln(\frac{n-1}{n})+ln(\frac{n+1}{n})
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  3. #3
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    Quote Originally Posted by General View Post
    1-
    \color{blue}{\pi}^n=(\sqrt{\pi})^{2n}
    \color{blue}9^n=3^{2n}
    Now, think about a well-known maclaurin series.

    2-
    Hint:
    \color{blue}ln(1-\frac{1}{n^2})=ln(\frac{n^2-1}{n^2})=ln(\frac{(n-1)(n+1)}{(n)(n)})=ln(\frac{n-1}{n})+ln(\frac{n+1}{n})
    Thank you mr blue
    the first is easy.
    the second i will break it to two series each one is telescoping
    Nyahahahaaaa
    Am math professor

    Thanks mr blue again
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  4. #4
    MHF Contributor chisigma's Avatar
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    The 'infinite sum'...

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

    ... is , in my opinion of course, 'very nice'. If we remember the 'infinite product'...

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

    ... it is easy to derive that is...

    \prod_{n=2}^{\infty} (1-\frac{1}{n^{2}}) = \lim_{x \rightarrow 1} \frac{sinc (x)}{1-x^{2}} = \frac{1}{2} (3)

    ... so that...

    \sum_{n=2}^{\infty} \ln (1-\frac{1}{n^{2}})= -\ln 2 (4)

    Kind regards

    \chi \sigma
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