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Math Help - series

  1. #1
    Super Member Random Variable's Avatar
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    series

    Challenge Problem:

    Find the sum of  \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7 \cdot 8} + \frac{1}{9 \cdot 10 \cdot 11 \cdot 12} + ...

    Moderator editor: Approved Challenge question.
    Last edited by CaptainBlack; May 16th 2010 at 08:13 PM.
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  2. #2
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    I am not sure my calculations are correct , please check it

    We have

      \prod_{k=0}^{n} ( x + k)^{-1} = \frac{1}{n!} \sum_{k=0}^n \frac{(-1)^k \binom{n}{k}}{ x + k}

    Therefore ,

      \frac{1}{ x(x+1)(x+2)(x+3) } = \frac{1}{6} \left( \frac{1}{x} - \frac{3}{x+1} + \frac{3}{x+2} - \frac{1}{x+3} \right )

    your series should be equal to


     \frac{1}{6} ( \frac{1}{1} - \frac{3}{2} + \frac{3}{3} - \frac{1}{4} +    \frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8} +  \frac{1}{9} - \frac{3}{10} + \frac{3}{11} - \frac{1}{12} + ...

     = \frac{1}{6} [ ( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... ) - 2 ( \frac{1}{2} - \frac{1}{3} + \frac{1}{6} - \frac{1}{7} + .... )]

     = \frac{1}{6} [ \ln{2} - 2 ( \int_0^1 ( x - x^2 + x^5 - x^6 + ... )~dx]

     = \frac{1}{6} \left[ \ln{2} - 2 \left( \int_0^1 [x( 1 + x^4 + ...) - x^2 ( 1 + x^4+...)]~dx \right ) \right ]

     = \frac{1}{6} [ \ln{2} - 2 \int_0^1 \frac{x(1-x)}{1 - x^4}~dx ]

    The integral
      \int_0^1 \frac{x(1-x)}{1 - x^4}~dx

     = \frac{1}{2}  \int_0^1 \left( \frac{x+1}{x^2+1} - \frac{1}{x+1} \right)~dx

     = \frac{1}{2} ( \frac{\pi}{4} - \frac{\ln{2}}{2} )

    The series

     = \frac{1}{6} [ \ln{2} - \frac{\pi}{4} +  \frac{\ln{2}}{2} ]

     = \frac{1}{6} [ \frac{3}{2} \ln{2}  - \frac{\pi}{4} ]

     = \frac{1}{24} ( 6\ln{2} - \pi )
    Last edited by simplependulum; May 16th 2010 at 09:03 PM.
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  3. #3
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Random Variable View Post
    Challenge Problem:

    Find the sum of  \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7 \cdot 8} + \frac{1}{9 \cdot 10 \cdot 11 \cdot 12} + ...

    Moderator editor: Approved Challenge question.
    This is \sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)}. Let us solve a more interesting sum.

    Let S_\ell=\sum_{n=1}^{\infty}\frac{1}{f_\ell(n)} where f_\ell(n)=\prod_{j=0}^{\ell}(n+j),\text{ }\ell\geqslant 1.

    Notice then that S_\ell=\sum_{n=1}^{\infty}\frac{1}{f_\ell(n)}=\sum  _{n=0}^{\infty}\frac{n!\ell!}{\ell!(n+\ell+1)!}.

    So, remembering our definitions this may be written as

    S_\ell=\sum_{n=0}^{\infty}\frac{\Gamma(n+1)\Gamma(  \ell+1)}{\ell!\Gamma(n+\ell+2)}=\frac{1}{\ell!}\su  m_{n=0}^{\infty}B(n+1,\ell+1).

    Thus, \ell! S_\ell=\sum_{n=0}^{\infty}\int_0^1 t^n(1-t)^\ell dt=\int_0^1 (1-t)^\ell\sum_{n=0}^{\infty}t^n dt

    And remembering the formula for a geometeric sum we have that

    \ell! S_\ell=\int_0^1(1-t)^{\ell-1}=\frac{1}{\ell}.


    Thus, S_\ell=\frac{1}{\ell \ell!}.

    Thus, S_3=\frac{1}{18}
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  4. #4
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by Drexel28 View Post
    This is \sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)(n+3)}.
    No, it's not!
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  5. #5
    MHF Contributor Bruno J.'s Avatar
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    Quote Originally Posted by simplependulum View Post
    I am not sure my calculations are correct , please check it



    your series should be equal to

     \frac{1}{6} ( \frac{1}{1} - \frac{3}{2} + \frac{3}{3} - \frac{1}{4} +    \frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8} +  \frac{1}{9} - \frac{3}{10} + \frac{3}{11} - \frac{1}{12} + ...

     = \frac{1}{6} [ ( \frac{1}{1} - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ... ) - 2 ( \frac{1}{2} - \frac{1}{3} + \frac{1}{6} - \frac{1}{7} + .... )]

     = \frac{1}{6} [ \ln{2} - 2 ( \int_0^1 ( x - x^2 + x^5 - x^6 + ... )~dx]

     = \frac{1}{6} \left[ \ln{2} - 2 \left( \int_0^1 [x( 1 + x^4 + ...) - x^2 ( 1 + x^4+...)]~dx \right ) \right ]

     = \frac{1}{6} [ \ln{2} - 2 \int_0^1 \frac{x(1-x)}{1 - x^4}~dx ]

    The integral
      \int_0^1 \frac{x(1-x)}{1 - x^4}~dx

     = \frac{1}{2}  \int_0^1 \left( \frac{x+1}{x^2+1} - \frac{1}{x+1} \right)~dx

     = \frac{1}{2} ( \frac{\pi}{4} - \frac{\ln{2}}{2} )

    The series

     = \frac{1}{6} [ \ln{2} - \frac{\pi}{4} +  \frac{\ln{2}}{2} ]

     = \frac{1}{6} [ \frac{3}{2} \ln{2}  - \frac{\pi}{4} ]

     = \frac{1}{24} ( 6\ln{2} - \pi )
    Mathematica agrees! Good job! Maybe the first step could be written more clearly (it's not obvious you used partial fractions!) but you did an impressive job there!
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by Bruno J. View Post
    No, it's not!
    Haha, wow that was a little mistake. I can remedy it anyways...

    So, let S=\sum_{n=0}^{\infty}\frac{1}{(4n+1)(4n+2)(4n+3)(4  n+4)}

    Using the exact same trick as my early proof we can transform this into

    S=\sum_{n=0}^{\infty}\frac{(4n+5)4!(4n!)}{4!(4n+5)  !}=\frac{1}{4!}\sum_{n=0}^{\infty}(4n+5)B(4n+1,4+1  )

    So,

    4!\cdot S=\sum_{n=0}^{\infty}(4n+5)\int_0^1 t^{4n}(1-t)^4 dt

    Switching integral and summation we get

    4!\cdot S=\int_0^1 (1-t)^4\sum_{n=0}^{\infty}(4n+5)t^{4n}=\int_0^1\frac{  (1-t)^4(5-t^4)}{(1-t^4)^2}dt\overset{\color{red}(*)}{=}6\ln(2)-\pi

    Where the star indicates integration by method of partial fractions.

    So, S=\frac{1}{24}\left(6\ln(2)-\pi\right)
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  7. #7
    MHF Contributor Drexel28's Avatar
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    Just for generalizations sake...

    Let S(\alpha,\beta)=\sum_{n=0}^{\infty}\frac{1}{(\alph  a n+1)\cdots(\alpha n+\beta)},\text{ }\alpha,\beta\in\mathbb{N},\text{ }\beta\geqslant 1

    Then, using the exact same method

    S(\alpha,\beta)=\frac{1}{\beta!}\sum_{n=0}^{\infty  }(\alpha n+\beta+1)B(\alpha n +1,\beta +1)

    Or, \beta!S(\alpha,\beta)=\int_0^1(1-t)^\beta\sum_{n=0}^{\infty}(\alpha n+\beta)t^{\alpha n}

    Or, \beta !S(\alpha,\beta)=\int_0^1\frac{(1-t)^\beta\cdot(\alpha t^\alpha+\beta(1-t^\alpha))}{(1-t^\alpha)^2}dt

    Wonder if anything meaningful lies in that beast
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  8. #8
    Super Member Random Variable's Avatar
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    I expressed each term as a quadruple integral, reversed the order of integration, and then summmed up the terms.


     \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)} = \int_{0}^{1} \int^{z}_{0}  \int^{y}_{0} \int^{x}_{0} w^{4n} \ dw \ dx \ dy \ dz

     = \int_{0}^{1} \int^{1}_{w} \int^{1}_{x} \int^{1}_{y} w^{4n} \ dz \ dy \   dx \ dw = \int^{1}_{0} w^{4n} \Big(\frac{1}{6} - \frac{w}{2} +  \frac{w^{2}}{2} - \frac{w^{3}}{6} \Big) \ dw =  = \frac{1}{6} \int^{1}_{0} w^{4n} (1-w)^{3} \ dw


     \sum^{\infty}_{n=0} \frac{1}{(4n+1)(4n+2)(4n+3)(4n+4)} = \frac{1}{6}  \sum^{\infty}_{n=0} \int^{1}_{0} w^{4n}(1-w)^{3} \ dw

     = \frac{1}{6}\int^{1}_{0} (1-w)^{3} \sum^{\infty}_{n=0} w^{4n}\  dw =  \frac{1}{6} \int^{1}_{0} \frac{(1-w)^{3}}{1-w^{4}} \ dw

     = \frac{1}{6} \int^{1}_{0} \frac{(1-w)^{3}}{(1-w)(1+w)(1+w^{2})} \ dw =  \frac{1}{6} \int^{1}_{0} \frac{(1-w)^{2}}{(1+w)(1+w^{2})} \ dw

      = \frac{1}{6} \int^{1}_{0} \Big(\frac{2}{1+w} - \frac{1}{1+w^{2}} -  \frac{w}{1+w^{2}} \ \Big) \ dw = \frac{1}{6} \Big( 2 \ln (1+w) - \arctan  w - \frac{\ln(1+w^{2})}{2} \Big) \Big|^{1}_{0}

     = \frac{1}{6} \Big( 2 \ln 2 - \frac{\pi}{4} - \frac{\ln 2}{2} \Big) = \frac{\ln 2}{4} - \frac{\pi}{24}
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  9. #9
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    Hi

    I used another method, maybe less tricky

    \frac{1}{1 \cdot 2 \cdot 3 \cdot 4} + \frac{1}{5 \cdot 6 \cdot 7 \cdot 8} + \frac{1}{9 \cdot 10 \cdot 11 \cdot 12} + ...  = \sum_{k=0}^{+\infty} \frac{(4k)!}{(4k+4)!}

    Let

    f(x) = \sum_{k=0}^{+\infty} \frac{(4k)!}{(4k+4)!} x^{4k+4}

    The sum is equal to f(1)

    The fourth derivative of f is f^{(4)}(x) = \sum_{k=0}^{+\infty}  x^{4k} = \frac{1}{1-x^4}

    The fourth anti-derivative of f^{(4)}(x) is

    f(x) = \frac{(1+x)^3}{24} \ln(1+x) +\frac{(1-x)^3}{24} \ln(1-x) - \frac{3x^2-1}{24} \ln(1+x^2) + \frac{x^3-3x}{12} \arctan(x)


    And f(1) = \frac{\ln(2)}{4} - \frac{\pi}{24}

    (using the fact that (1-x)^3 \ln(1-x) has 0 as limit when x approaches 1)
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  10. #10
    Super Member Random Variable's Avatar
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    How did you go from the first line to the second line? My brain isn't working.
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  11. #11
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    Quote Originally Posted by Random Variable View Post
    How did you go from the first line to the second line? My brain isn't working.

    I have put 3 = 1 + 2
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