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Math Help - Sum of the series.

  1. #1
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    Sum of the series.

    Hi! I need help in finding the sum of the series:

    I would appreciate any kind of help here, since I don't even know how to start.
    Last edited by Rist; June 10th 2008 at 01:48 PM. Reason: Trying to fix image
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Rist View Post
    Hi! I need help in finding the sum of the series:

    I would appreciate any kind of help here, since I don't even know how to start.
    I would say your image isn't loading.

    -Dan
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  3. #3
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Rist View Post
    Hi! I need help in finding the sum of the series:

    I would appreciate any kind of help here, since I don't even know how to start.
    For the first one, both x=0 and x=1 give divergent series

    as for the second one I am not sure that can be easily summed
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  4. #4
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    The second is easily summed: x = 12\quad  \Rightarrow \quad \sum\limits_{n = 1}^\infty  {\frac{1}{{n\left( {n + 1} \right)2^n }}}
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  5. #5
    Super Member wingless's Avatar
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    I think that x=0,1 means x=1/10 which gives a convergent series. Anyway I'll solve it for general.

    S = \sum_{n=0}^{\infty} \frac{x^n}{n+1}

    x\cdot S = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1}

    x\cdot S = \int \sum_{n=0}^{\infty} \frac{\partial}{\partial x} \frac{x^{n+1}}{n+1}~dx

    x\cdot S = \int \sum_{n=0}^{\infty} x^n~dx

    x\cdot S = \int \frac{1}{1-x} ~dx

    x\cdot S = - \ln (1-x) + C

    We can find C using,

    x\cdot S = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} = - \ln (1-x) + C

    Use x = 0, then
    0=-\ln (1-0)+C

    C=0

    So finally,

    x\cdot S = - \ln (1-x)

    S = - \frac{\ln (1-x)}{x}

    \sum_{n=0}^{\infty} \frac{x^n}{n+1} = - \frac{\ln (1-x)}{x}

    ---

    Assuming x=0.1 (this is the x in the original question, so 1-6x makes 0.4),

    S = - \frac{\ln (1-0.4)}{0.4} \approx 1.27706
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  6. #6
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    \sum\limits_{n\,=\,1}^{\infty }{\frac{1}{2^{n}n(n+1)}}=\sum\limits_{n\,=\,1}^{\i  nfty }{\frac{(n-1)!}{2^{n}(n+1)!}}=\sum\limits_{n\,=\,1}^{\infty }{\frac{\Gamma (n)\Gamma (2)}{2^{n}\Gamma (n+2)}}.

    Hence the sum equals,

    \sum\limits_{n\,=\,1}^{\infty }{\int_{0}^{1}{\frac{x^{n-1}(1-x)}{2^{n}}\,dx}}=\int_{0}^{1}{\left\{ \frac{1-x}{x}\sum\limits_{n\,=\,1}^{\infty }{\left( \frac{x}{2} \right)^{n}} \right\}\,dx}=\int_{0}^{1}{\frac{1-x}{2-x}\,dx}

    and the conclusion follows. (This sum can be tackled from many ways.)
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  7. #7
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Plato View Post
    The second is easily summed: x = 12\quad \Rightarrow \quad \sum\limits_{n = 1}^\infty {\frac{1}{{n\left( {n + 1} \right)2^n }}}
    Quote Originally Posted by Krizalid View Post
    \sum\limits_{n\,=\,1}^{\infty }{\frac{1}{2^{n}n(n+1)}}=\sum\limits_{n\,=\,1}^{\i  nfty }{\frac{(n-1)!}{2^{n}(n+1)!}}=\sum\limits_{n\,=\,1}^{\infty }{\frac{\Gamma (n)\Gamma (2)}{2^{n}\Gamma (n+2)}}.

    Hence the sum equals,

    \sum\limits_{n\,=\,1}^{\infty }{\int_{0}^{1}{\frac{x^{n-1}(1-x)}{2^{n}}\,dx}}=\int_{0}^{1}{\left\{ \frac{1-x}{x}\sum\limits_{n\,=\,1}^{\infty }{\left( \frac{x}{2} \right)^{n}} \right\}\,dx}=\int_{0}^{1}{\frac{1-x}{2-x}\,dx}

    and the conclusion follows. (This sum can be tackled from many ways.)
    That's what you call easily summed?
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  8. #8
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Mathstud28 View Post
    That's what you call easily summed?
    For Krizalid, that's pretty straightforward. He's not the "Inte-killer" for nothing!

    -Dan
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  9. #9
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    Quote Originally Posted by Mathstud28 View Post

    That's what you call easily summed?
    I don't say that this is called "easily summed." Why do you say that?
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    I don't say that this is called "easily summed." Why do you say that?
    Plato: The second is easily summed:

    ^^
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  11. #11
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    Quote Originally Posted by Mathstud28 View Post
    That's what you call easily summed?
    Since \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}:


    S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)2^n} = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^n}


     = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - 2 \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^{n+1}} ........
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by mr fantastic View Post
    Since \frac{1}{n(n+1)} = \frac{1}{n} - \frac{1}{n+1}:

    S = \sum_{n=1}^{\infty} \frac{1}{n(n+1)2^n} = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^n} = \sum_{n=1}^{\infty} \frac{1}{n \, 2^n} - 2 \sum_{n=1}^{\infty} \frac{1}{(n+1) \, 2^{n+1}} ........
    Wow haha...this has really turned into a big deal. I was stating that for most people this wouldn't be the easiest to sum series.
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  13. #13
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    Ahhh well, but, why are you talkin' to me for, if Plato said so.
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  14. #14
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Krizalid View Post
    Ahhh well, but, why are you talkin' to me for, if Plato said so.
    I quoted you for your solution so that when I said "That whats you call easy to sum" to Plato it had dramatic effect

    Note to self: Don't say things are easy
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  15. #15
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    *Ahem*
    Quote Originally Posted by Mathstud28 View Post
    [snip]
    as for the second one I am not sure that can be easily summed
    Plato remarked that it was.

    I doubt Krizalid's approach was what Plato had in mind (which in no way is meant to diminish K's solution).

    My post is what Plato might have had in mind (I won't presume that it was). I think it's evident that Plato's remark was both reasonable and accurate.


    Quote Originally Posted by Mathstud28 View Post
    Wow haha...this has really turned into a big deal. I was stating that for most people this wouldn't be the easiest to sum series.
    True, since most people don't study to the level that this question is contexted within.
    Last edited by mr fantastic; June 10th 2008 at 06:42 PM. Reason: Added the last paragraph
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