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Thread: Series of Binomial Coefficients

  1. #1
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    Jan 2007
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    Series of Binomial Coefficients

    I'm trying to solve an infinite series (it's part of a larger applied problem I'm working with, but I know what to do with this once I find it, so it's not relevant).

    I need the sum from n = k to n = infinity of

    C(n,k)*c^(-n)

    for a positive integer c > 1.

    It's a little different, sort of a reverse of the binomial theorem since the n runs with the series with k held constant.

    I have found that

    C(n+j,k) = C(n,k)*(n+j)!(n-k)!/[n!(n+j-k)!]

    But that seemed to cause more harm than good...

    I know from Mathematica the sum is going to be c/(c-1)^(k+1), but I'm having lots of trouble proving this.

    Thanks in advance for any assistance.
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  2. #2
    Member
    Joined
    Nov 2009
    Posts
    106
    We want the coefficient of $\displaystyle y^k$ is the following (then put $\displaystyle x=c^{-1}$):

    $\displaystyle \displaystyle{
    \sum_{m=0}^\infty y^m\sum_{n=0}^\infty \binom{n}{m} x^n=\sum_{n=0}^\infty \sum_{m=0}^\infty \binom{n}{m} y^mx^n
    }$

    $\displaystyle \displaystyle{
    =\sum_{n=0}^\infty (1+y)^nx^n = \frac{1}{1-x-xy}=\frac{1}{1-x}\cdot\frac{1}{1-\frac{x}{1-x}y}
    }$

    $\displaystyle \displaystyle{
    =\sum_{m=0}^\infty \frac{x^m}{(1-x)^{m+1}}y^m
    }$

    Therefore

    $\displaystyle \displaystyle\sum_{n=0}^\infty \binom{n}{k}c^{-n}=\frac{c^{-k}}{(1-c^{-1})^{k+1}}=\frac{c}{(c-1)^{k+1}}$



    Another way: let

    $\displaystyle \displaystyle{
    S_k=\sum_{n=0}^\infty x^n\binom{n}{k}
    }$

    Then, using $\displaystyle \binom{n}{k}=\binom{n+1}{k+1}-\binom{n}{k+1}$ which follows from Pascal's identity, we get

    $\displaystyle \displaystyle{S_k=\left(\frac{1}{x}-1}\right)S_{k+1}}$

    or equivalently:

    $\displaystyle \displaystyle{
    S_{k+1} = \frac{x}{1-x}S_k
    }$

    and therefore

    $\displaystyle
    \displaystyle{S_{k}=\frac{x^{k-1}}{(1-x)^{k-1}}\sum_{n=0}^\infty \binom{n}{1}x^n}=
    \frac{x^{k-1}}{(1-x)^{k-1}}\sum_{n=0}^\infty nx^n}
    $

    $\displaystyle \displaystyle{=\frac{x^{k-1}}{(1-x)^{k-1}}\frac{x}{(1-x)^2}
    $

    As we got earlier.

    Edit:
    Actually, in the second method, it would be easier to go down to $\displaystyle S_0$ instead of $\displaystyle S_1$:

    $\displaystyle
    \displaystyle{S_{k}=\frac{x^k}{(1-x)^k}S_0 = \frac{x^k}{(1-x)^k}\cdot\frac{1}{1-x}
    $
    Last edited by Unbeatable0; Feb 6th 2011 at 04:51 AM.
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