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A Finite series and product of Gammas

  1. Apr 28, 2017 #1

    ShayanJ

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    Gold Member

    In reviewing some calculations, I've arrived at the series:

    ##S(d)=-\frac 1 {d-1}+\frac 1 2 \frac{d-2}{d-3}-\frac 1 8 \frac{(d-2)(d-4)}{d-5}+\frac 1 {48} \frac{(d-2)(d-4)(d-6)}{d-7}+\dots ##

    Its an infinite series but because I'm interested in its values for even ##d##s, its actually a finite series at each evaluation. Its values are:

    ## \begin{array}{c|ccccc} d \ \ \ &2 \ \ \ &4 \ \ \ &6 \ \ \ &8 \ \ \ &10 \ \ \ &12 \\\hline S(d) \ \ \ &-1 \ \ \ &\frac 2 3 \ \ \ &-\frac 8 {15} \ \ \ &\frac{16}{35} \ \ \ &-\frac{128}{315} \ \ \ &\frac{256}{693}\end{array} ##

    But the interesting fact is that the expression ## G(d)=\frac{\Gamma(\frac d 2)\Gamma(\frac{1-d}2)}{2 \sqrt \pi} ## has the exact same values for at least the above values of ##d##. So it seems that I should be able to prove that ##S(d)=G(d)##. But I have no idea how to do it. At first I tried to somehow write the series in terms of the Hypergeometric function, but that is both impossible and useless. So, any ideas?

    EDIT: Another evidence is that both ##S(d)## and ##G(d)## are singular for odd ##d##s.

    Thanks
     
    Last edited: Apr 28, 2017
  2. jcsd
  3. Apr 28, 2017 #2

    fresh_42

    Staff: Mentor

    Just for all who like to solve this riddle like me, the statement is equivalent to:
    $$
    - \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \cdot \frac{1}{2n-(2k+1)}\cdot \prod_{i=1}^{k}(n-i) = \frac{n!(n-1)!}{2(2n)!}\cdot (-4)^n = (-1)^n \cdot \frac{2^{2n}}{(2n)\binom{2n}{n}}
    $$
     
    Last edited: Apr 28, 2017
  4. Apr 28, 2017 #3

    mfb

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    2016 Award

    Staff: Mentor

    The denominators also appear in the series expansion of ##(1-x)^{-3/2}##.

    OEIS sequence. We even get the numerators there with this interesting comment:
     
  5. Apr 28, 2017 #4

    ShayanJ

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    Gold Member

    It should be ...(n-2i)...

    Anyway, are you guys saying that this is a famous unsolved problem?
     
  6. Apr 28, 2017 #5

    mfb

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    2016 Award

    Staff: Mentor

    Can you give more terms?
    Who is saying that?
     
  7. Apr 28, 2017 #6

    fresh_42

    Staff: Mentor

    No. The factor ##2## has been canceled by setting ##d=2n##. (It's also missing in ##2 \cdot 4 \cdot \ldots \cdot 2k##)
    I took the formulas from the German Wiki page of the Gamma function.
     
  8. Apr 28, 2017 #7

    ShayanJ

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    Gold Member

    Never mind, he clarified.

    I just got that impression, specially because of your post since I'm not quite sure how you could find that information without previous knowledge of the problem.
     
  9. Apr 28, 2017 #8

    mfb

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    Staff: Mentor

    I searched for the first denominators on OEIS, then calculated more using fresh's formula to find the right sequence, and confirmed it via the given formulas there.
     
  10. Apr 29, 2017 #9

    fresh_42

    Staff: Mentor

    I don't know, whether it's already obvious to you, but in any case and for the records: with an induction argument (or simply a recursion on the right hand side), it boils down to show
    $$ - \sum_{k=0}^{\infty} \frac{(-1)^{k}}{k!} \cdot \frac{1}{d-(2k+1)}\cdot \prod_{i=1}^{k}(\frac{d}{2}-i) = (-1)^{\frac{d}{2}} \cdot \frac{2}{3} \cdot \frac{4}{5} \cdot \frac{6}{7} \cdot \ldots \cdot \frac{d-2}{d-1}
    $$
    which itself can very likely be shown by another induction.
     
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