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Math Help - Ad hoc summation of series

  1. #1
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    Ad hoc summation of series

    Hello,

    is there a way of ad hoc summing 1/(1*2*3*4) + 1/(5*6*7*8) + ... without getting into the theory of Polygamma functions?

    Thanks
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  2. #2
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    You could look at the telescopic series

    \displaystyle \sum_{n=1}^{\infty} \dfrac{1}{n(n+1)(n+2)(n+3)} = \dfrac{1}{6}\sum_{n=1}^{\infty} \left(\dfrac{1}{n} - \dfrac{1}{n+3}\right) - \dfrac{1}{2} \sum_{n=1}^{\infty} \left(\dfrac{1}{n+1} - \dfrac{1}{n+2}\right)

    Edit: Sorry wrong series
    Last edited by Jester; January 2nd 2011 at 08:19 AM. Reason: wrong series
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  3. #3
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    Hello, Heirot!

    is there a way of ad hoc summing: . S \;=\;\dfrac{1}{1\cdot2\cdot3\cdot4} + \dfrac{1}{5\cdot6\cdot7\cdot8} + \hdots

    without getting into the theory of Polygamma functions?

    The general term is: . a_n \;=\;\dfrac{1}{n(n+1)(n+2)(n+3)}


    Partial fractions: . \displaystyle \frac{1}{n(n+1)(n+2)(n+3)} \;=\;\frac{A}{n} + \frac{B}{n+1} + \frac{C}{n+2} + \frac{D}{n+3}

    We find that: . A = \frac{1}{6},\;B = -\frac{1}{2},\;C = \frac{1}{2},\;D = -\frac{1}{6}

    Hence: . \displaystyle a_n \;=\;\frac{\frac{1}{6}}{n} - \frac{\frac{1}{2}}{n+1} + \frac{\frac{1}{2}}{n+2} - \frac{\frac{1}{6}}{n+3} \;=\;\frac{1}{6}\left(\frac{1}{n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)

    Then: . \displaystyle S \;=\;\frac{1}{6}\sum^{\infty}_{n=1}\left(\frac{1}{  n} - \frac{3}{n+1} + \frac{3}{n+2} - \frac{1}{n+3}\right)


    Crank out several terms:

    \displaystyle S \;=\;\frac{1}{6}\bigg[\left(1 - \frac{3}{2} + \frac{3}{3} - \frac{1}{4}\right) + \left(\frac{1}{2} - \frac{3}{3} + \frac{3}{4} - \frac{1}{5}\right) + \left(\frac{1}{3} - \frac{3}{4} + \frac{3}{5} - \frac{1}{6}\right)

    . . . . . . . \displaystyle + \left(\frac{1}{4} - \frac{3}{5} + \frac{3}{6} - \frac{1}{7}\right) + \left(\frac{1}{5} - \frac{3}{6} + \frac{3}{7} - \frac{1}{8}\right) + \left(\frac{1}{6} - \frac{3}{7} + \frac{3}{8} - \frac{1}{9}\right)

    . . . . . . . . . \displaystyle + \left(\frac{1}{7} - \frac{3}{8} + \frac{3}{9} - \frac{1}{10}\right) + \left(\frac{1}{8} - \frac{3}{9} + \frac{3}{10} - \frac{1}{11}\right) + \hdots \bigg]


    We find that nearly all the terms cancel out.

    And we are left with: . \displaystyle S \;=\;\frac{1}{6}\left(1 - \frac{3}{2} + \frac{1}{2} + \frac{1}{3}\right) \;=\;\frac{1}{18}

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  4. #4
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    The general term is in fact a_n = 1 / (4n (4n-1) (4n-2) (4n-3)) = (4(n-1))! /(4n)! and this it what complicates the sum. But thanks for trying, guys!
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  5. #5
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Heirot View Post
    ... But thanks for trying, guys!
    What do you mean by that? They both are correct.
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  6. #6
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    Quote Originally Posted by chiph588@ View Post
    What do you mean by that? They both are correct.
    \displaystyle\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+  2)(n+3)}=\frac{1}{1(2)3(4)}+\frac{1}{2(3)4(5)}+\fr  ac{1}{3(4)5(6)}+....
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  7. #7
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    IF you start with the geometric power series

    \displaystyle \dfrac{1}{1-x} = \sum_{n=0}^{\infty} x^n
    then replace  x with x^4 then

    \displaystyle \dfrac{1}{1-x^4} = \sum_{n=0}^{\infty} x^{4n}

    integrate 4 times noting that the constants of integration are found by setting x = 0 into both sides of the equation. Then taking the limit as  x \to 1 gives your answer

    Spoiler:
    \text{I got }\;S = \dfrac{1}{4} \ln 2 - \dfrac{\pi}{24}
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  8. #8
    MHF Contributor chiph588@'s Avatar
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    \begin{aligned} \displaystyle \sum_{n=1}^\infty \frac1{4n(4n+1)(4n+2)(4n+3)} &= \sum_{n=1}^\infty\left(\frac{1/6}{4n}-\frac{1/2}{4n+1}+\frac{1/2}{4n+2}-\frac{1/6}{4n+3}\right)\\<br />
&= \int_0^1\sum_{n=1}^\infty \left(\frac16x^{4n-1} -\frac12x^{4n}+\frac12x^{4n+1}-\frac16x^{4n-2}\right) dx\\<br />
&= \int_0^1\left(-\frac16\frac{x^3}{x^4-1}+\frac12\frac{x^4}{x^4-1}-\frac12\frac{x^5}{x^4-1}+\frac16\frac{x^6}{x^4-1}\right)dx\\<br />
&=  -\frac16\int_0^1\left(\frac{x^3}{x^4-1}-\frac1{4(x-1)}\right)dx+\frac12\int_0^1\left(\frac{x^4}{x^4-1}-\frac1{4(x-1)}\right)dx\\&-\frac12\int_0^1\left(\frac{x^5}{x^4-1}-\frac1{4(x-1)}\right)dx+\frac16\int_0^1\left(\frac{x^6}{x^4-1}-\frac1{4(x-1)}\right)dx\\<br />
&=-\frac1{12}\log2 +\frac12\left(1-\frac\pi8-\frac{\log2}4\right)-\frac14+\frac16\left(\frac1{24}(8+3\pi-6\log2)\right)\\<br />
&= \frac1{72}(22-3\pi-18\log2) <br />
 \end{aligned}

    Note, those four integrals are a pain though...

    Edit: I did the wrong series! Your answer can be derived very similarly though..
    Last edited by chiph588@; January 2nd 2011 at 12:18 PM.
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  9. #9
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    You might need another series or a shift.
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  10. #10
    MHF Contributor Also sprach Zarathustra's Avatar
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    \frac{1}{4n(4n-1)(4n-2)(4n-3)}=\frac{1}{8}(\frac{4}{3(4n-3)}+\frac{4}{4n-1}-\frac{2}{2n-1}-\frac{1}{3n})
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  11. #11
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    We may write S = \displaystyle\sum_{n\geq 0}{\frac{1}{(4\cdot n + 1)...(4\cdot n + 4)}} = \frac{1}{6}\cdot \sum_{n\geq 0}{\text{B}(4n+1,4)} where \text{B}(x,y) is the beta function.

    Thus 6\cdot S = \displaystyle\sum_{n\geq 0}{\int_0^1{x^{4n}\cdot (1-x)^3}dx} = \int_0^1\left(\sum_{n\geq 0}x^{4n}\right)\cdot (1-x)^3dx = \int_0^1{\frac{(1-x)^3}{1-x^4}dx}

    Expanding the numerator: 6\cdot S = \displaystyle\int_0^1{\frac{(1-x)^2}{(1+x)\cdot (1+x^2)}dx}= \log(2) - \int_0^1{\frac{2x}{(1+x^2)}\cdot \frac{dx}{1+x}}

    Note that : \displaystyle\frac{2x}{(1+x)\cdot (1+x^2)} = \displaystyle\frac{1+x}{1+x^2} - \frac{1}{1+x}

    And integrate.
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  12. #12
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    *ignore this post... I misread the summation the OP posted*
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  13. #13
    MHF Contributor FernandoRevilla's Avatar
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    An alternative. Let:

    S=\displaystyle\sum_{n=1}^{+\infty}a_n

    the sum of the given series. Using Soroban's decomposition:

    a_n=b_n+c_n,\quad b_n=\dfrac{1}{6}\left(\dfrac{1}{n}-\dfrac{1}{n+3}\right),\quad c_n=\dfrac{1}{2}\left(\dfrac{1}{n+2}-\dfrac{1}{n+1}\right)

    Now we use the well known property:

    \displaystyle\sum_{n=1}^{+\infty}(\varphi(n+q)-\varphi(n))=q\displaystyle\lim_{n \to{+}\infty}{\varphi (n)}-[ \varphi(1)+ \varphi (2)+\ldots + \varphi (q)]

    we inmediately obtain:

    S_1=\displaystyle\sum_{n=1}^{+\infty}b_n=\dfrac{11  }{36},\quad S_2=\displaystyle\sum_{n=1}^{+\infty}c_n=-\dfrac{1}{4}

    The sum of the given series is:

    S=S_1+S_2=\dfrac{11}{36}-\dfrac{1}{4}=\dfrac{1}{18}

    Fernando Revilla

    Edited: What a strange thread! . I used Soroban's series instead of OP's series.
    Last edited by FernandoRevilla; January 2nd 2011 at 01:25 PM.
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  14. #14
    MHF Contributor Also sprach Zarathustra's Avatar
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    Why a_n=B(4n+1,4) ?
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  15. #15
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by Also sprach Zarathustra View Post
    Why a_n=B(4n+1,4) ?
    It's because  \text{B}(x,y) = \frac{\Gamma(x)\Gamma(y)}{\Gamma(x+y)} .

    See here.
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