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Hypergeometric function. Summation question

  1. Dec 1, 2015 #1
    1. The problem statement, all variables and given/known data
    It is very well known that ## \sum^{\infty}_{n=0}x^n=\frac{1}{1-x}##. How to show that
    ## \sum^{\infty}_{n=0}\frac{(a)_n}{n!}x^n=\frac{1}{(1-x)^a}##
    Where ##(a)_n=\frac{\Gamma(a+n)}{\Gamma(a)}##



    2. Relevant equations
    ## \Gamma(x)=\int^{\infty}_0 e^{-t}t^{x-1}dt##

    3. The attempt at a solution
    ## \frac{(a)_n}{n!}=\frac{a(a+1)...(a+n-1)}{n!}##
    Not sure how to go further.
     
  2. jcsd
  3. Dec 1, 2015 #2
    Let's call for ##\alpha < 0## : ##f_\alpha(x)= (1-x)^\alpha## and ## R_{n,\alpha}(x) = f_\alpha(x) - \sum_{k=0}^n \frac{f_\alpha ^ {(k)} (0)}{k!} x^k ##

    Can you explain why ##|x| < 1 ## and ##x## in a neighborhood of 0 implies ##R_{n,\alpha}(x) = \int_0^x \frac{(x-t)^n}{n!} f_\alpha ^{(n+1)}(t) \ dt##. Then show that ##\lim_{n\to \infty} R_{n,\alpha}(x) = 0 ##
     
  4. Dec 1, 2015 #3

    Ray Vickson

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    Science Advisor
    Homework Helper

    Just evaluate and simplify the ##n##th term of the Maclaurin expansion of ##f(x) = (1-x)^{-a}##.

    Basically, you are being asked to verify that the binomial expansion of ##(1-x)^n## applies to non-integer and/or negative values of ##n##. However, you cannot use the "factorial" definition of the binomial coefficient ##C^n_k## when ##n## is not a positive integer; instead, you need to use the explicit definition
    [tex] C^n_k = \frac{n (n-1) \cdots (n-k+1)}{k!} [/tex]
    for integers ##k \geq 0##.
     
  5. Dec 1, 2015 #4
    Thx a lot.
     
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