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Math Help - having a problem in determine the Convergence or divergence for infinite sum.

  1. #1
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    having a problem in determine the Convergence or divergence for infinite sum.

    can anyone show me how to
    Determine whether ∑(k=1 to n) [(1-ksin(1/k)]
    converges .
    TQ
    Last edited by younhock; March 2nd 2010 at 12:48 AM.
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  2. #2
    MHF Contributor chisigma's Avatar
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    If the general term of the series is a_{k} = 1- \sin \frac{1}{k} the series diverges because is \lim_{k \rightarrow \infty} a_{k} = 1 \ne 0... may be that the a_{k} are something else?...

    Kind regards

    \chi \sigma
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  3. #3
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    Quote Originally Posted by chisigma View Post
    If the general term of the series is a_{k} = 1- \sin \frac{1}{k} the series diverges because is \lim_{k \rightarrow \infty} a_{k} = 1 \ne 0... may be that the a_{k} are something else?...

    Kind regards

    \chi \sigma

    oh yea.. mistype a k there.. updated.. tQ.
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  4. #4
    MHF Contributor chisigma's Avatar
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    All right!... so the series is...

    \sum_{k=1}^{\infty} (1-k\cdot \sin \frac{1}{k}) (1)

    Using the series expansion...

    \sin x= \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot x^{2n+1}}{(2n+1)!} (2)

    ... we find that...

    1 - k\cdot \sin \frac{1}{k} = - \sum_{n=1}^{\infty}\frac{(-1)^{n}}{k^{2n}\cdot (2n+1)!} (3)

    ... so that the (1) becomes...

    \sum_{k=1}^{\infty} (1-k\cdot \sin \frac{1}{k})= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{k^{2n}\cdot (2n+1)!}=

    = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cdot \zeta (2n)}{(2n+1)!} < 2\cdot \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n+1)!} = 2\cdot (1-\sin 1) (4)

    ... and the series (1) converges...

    Kind regards

    \chi \sigma
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  5. #5
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    Quote Originally Posted by chisigma View Post
    All right!... so the series is...

    \sum_{k=1}^{\infty} (1-k\cdot \sin \frac{1}{k}) (1)

    Using the series expansion...

    \sin x= \sum_{n=0}^{\infty} \frac{(-1)^{n}\cdot x^{2n+1}}{(2n+1)!} (2)

    ... we find that...

    1 - k\cdot \sin \frac{1}{k} = - \sum_{n=1}^{\infty}\frac{(-1)^{n}}{k^{2n}\cdot (2n+1)!} (3)

    ... so that the (1) becomes...

    \sum_{k=1}^{\infty} (1-k\cdot \sin \frac{1}{k})= \sum_{k=1}^{\infty} \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{k^{2n}\cdot (2n+1)!}=

    = \sum_{n=1}^{\infty}\frac{(-1)^{n-1}\cdot \zeta (2n)}{(2n+1)!} < 2\cdot \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(2n+1)!} = 2\cdot (1-\sin 1) (4)

    ... and the series (1) converges...

    Kind regards

    \chi \sigma

    TQ.Just wanna share with u, i found another alternative way of solving this question and it is much more simpler .By using limit comparison test comparing to 1/k^2 .
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