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Math Help - Series equation

  1. #1
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    Series equation

    Could anyone help me out with this equation please?

    \sum_{n=2}^{\infty} \frac{2}{(n^{3} - n)3^{n}} = - \frac{1}{2} + \frac{4}{3} \sum_{n=1}^{\infty} \frac {1}{n3^{n}}

    How to prove it? Any ideas?
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  2. #2
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    Re: Series equation

    Yes consider the partial fraction expansion of \displaystyle \frac{1}{n^3-n}.

    Two of the fractions will telescope and you'll be done.

    P.S. join mathhelpboards.com - this one is done for.
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  3. #3
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    Re: Series equation

    Could you write it out please? You mean  \frac{1}{n(n-1)(n+1)} ?
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  4. #4
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    Re: Series equation

    Yes, try to write it as

    {A \over n} + {B \over n - 1} + {C \over n + 1}

    Deduce what A, B, and C are.
    Do you know how to?
    Then split the summation into 3 summations.
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  5. #5
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    Re: Series equation

    OK, I did it.  A = -1, B = \frac {1}{2}, C = \frac{1}{2} . Is it ok?
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    Re: Series equation

    Quote Originally Posted by Lisa91 View Post
    OK, I did it.  A = -1, B = \frac {1}{2}, C = \frac{1}{2} . Is it ok?
    Yes.
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    Re: Series equation

    So, I got something like this:  -\frac{2}{n3^{n}} + \frac{1}{(n-1)3^{n}} + \frac{1}{(n+1)3^{n}} I still don't see what I can take out of this...
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  8. #8
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    Re: Series equation

    You have a summation that starts at n=2.

    Rewrite the summation of the second term by replacing n-1 by for instance k.
    And rewrite the summation of the third term by replacing n+1 by m.

    Afterward add or subtract whatever you need to, to make the summations start at 1.
    Then you can combine them again.
    Last edited by ILikeSerena; January 15th 2013 at 11:09 AM.
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  9. #9
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    Re: Series equation

    Oh sorry stupid question.
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    Re: Series equation

    Quote Originally Posted by Lisa91 View Post
    Oh sorry stupid question.
    If you didn't understand that then write out the terms of the series and cancel what you can cancel.
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  11. #11
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    Re: Series equation

    You can split the summation into 3 summations.

    The second summation is:

    \sum_{n=2}^\infty \frac 1 {(n-1)3^n}

    Substitute n=k+1 and you'll get:

    \sum_{k+1=2}^\infty \frac 1 {(k+1-1)3^{k+1}} = \sum_{k=1}^\infty \frac 1 {k 3^{k+1}} = \frac 1 3 \sum_{n=1}^\infty \frac 1 {n 3^n}
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  12. #12
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    Re: Series equation

    The third term is \sum_{m=3}^{\infty} \frac{3}{m3^{m}} . How can I 'go back' to 1?
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    Re: Series equation

    By subtracting the terms at m = 2, and m = 1.

    \sum_{m=1}^{\infty} \frac{3}{m3^{m}} -\frac{3}{m3^{m}}\bigg|_{m=1}-\frac{3}{m3^{m}}\bigg|_{m=2}
    Last edited by TheSaviour; January 15th 2013 at 11:37 AM.
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  14. #14
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    Re: Series equation

    So, I got \sum_{m=1}^{\infty} \frac{3}{m3^{m}} - \frac{7}{6} + \frac{1}{3}\sum_{k=1}^{\infty} \frac{1}{k3^{k}} - \sum_{n=1}^{\infty} \frac{2}{n3^{n}} How to combine them all? I don't know how to cope with those  2,3, \frac{1}{3} . For instance, if I go back to n, I get n=0 in the first term, which is not what I want...
    Last edited by Lisa91; January 15th 2013 at 11:46 AM.
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  15. #15
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    Re: Series equation

    3S+S/3-2S Surely you can combine that.

    Make all indexes k if that's confusing you.
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