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Find the sum(1/4!)+(4!/8!)+(8!/12!)+

  1. Nov 20, 2008 #1
    find the sum
    (1/4!)+(4!/8!)+(8!/12!)+...
     
  2. jcsd
  3. Nov 20, 2008 #2
    Re: sum

    That is clearly divergent. Try to simplify it and you will see.
     
  4. Nov 20, 2008 #3

    mathman

    User Avatar
    Science Advisor
    Gold Member

    Re: sum

    Not so!!!

    An upper bound would be 1 + 1/5^4 + 1/9^4 + 1/13^4 + ... which converges.
     
  5. Nov 20, 2008 #4
    Re: sum

    Regardless, the OP has shown no work towards the problem.
     
  6. Nov 21, 2008 #5
    Re: sum

    I can't see the trick to solve it offhand. We know it converges, but do you know for sure it converges to something nice? This seems like a somewhat-hard problem because the number 4 is actually important.

    [tex] \sum _{n=0}^\infty \frac{(k n)!}{( k (n+1) )!}[/tex]

    See if k is 1 then it will be the harmonic series which diverges.
     
  7. Nov 21, 2008 #6
    Re: sum

    (1/4!)+(4!/8!)+(8!/12!)+(12!/16!)+...
     
  8. Nov 21, 2008 #7
    Re: sum

    the answer contains Pi and Ln .
     
  9. Nov 21, 2008 #8
    Re: sum

    Yes, according to Mathematica the answer is
    [tex]
    \frac{1}{24}(6\log2-\pi)\approx 0.0423871
    [/tex]
    Did you solve this manually?
     
  10. Nov 21, 2008 #9
    Re: sum

    (1/4!)+(1/8*7*6*5)+(1/12*11*10*9)+...
    we can guess the general sentence .after that use definite integrals.
    thats all.
     
  11. Nov 22, 2008 #10
    Re: sum

    [tex] \sum _{k=0}^\infty \frac{1}{(4k+1)(4k+2)(4k+3)(4k+4)}[/tex]
     
  12. Nov 22, 2008 #11
    Re: sum

    Yes, and using
    [tex]
    a^{-1}=\int_0^\infty{dxe^{-ax}}
    [/tex]
    this can be written as
    [tex]
    \sum_{k=0}^\infty{\int_0^\infty{dx_1e^{-x_1(4k+1)}}\int_0^\infty{dx_2e^{-x_2(4k+2)}}\int_0^\infty{dx_3e^{-x_3(4k+3)}}\int_0^\infty{dx_4e^{-x_4(4k+4)}}}=\int_{[0,\infty)^4}{d^4(x_1,x_2,x_3,x_4)\frac{e^{-x_1-2x_2-3x_3-4x_4}}{1-e^{-4(x_1+x_2+x_3+x_4)}}}
    [/tex]
    Letting [itex]y_n=e^{-nx_n}[/itex] the last expression becomes
    [tex]
    \frac{1}{24}\int_{[0,1]^4}{d^4(y_1,y_2,y_3,y_4)}\frac{1}{1-y_1^4y_2^2y_3^{3/4}y_4}
    [/tex]
    These integrals can be solved explicitly but it's not nice. Did you do it this way?
     
    Last edited: Nov 22, 2008
  13. Nov 22, 2008 #12
    Re: sum

    [tex]\sum_{k=0}^infty \frac{1} {(4k+1)(4k+2)(4k+3)(4k+4) [/tex]
    is equal to
    [tex]\sum(\frac{1} {6(4k+1)} - \frac{1} {2(2k+1)} + \frac{1} {2(4k+3)} - \frac{1} {6(4k+1)} )[/tex]
    and use integrals
     
  14. Nov 22, 2008 #13
    Re: sum

    [tex]\sum(\frac{1} {6(4k+1)} - \frac{1} {2(4k+1)} + \frac{1} {2(4k+3)} - \frac{1} {6(4k+1)} )[/tex]
     
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