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Math Help - Pseudo-Geometric Series

  1. #1
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    Pseudo-Geometric Series

    Find a way to obtain an explicit formula, in terms of r such that -1 < r < 1, for \sum_{n=1} ^{\infty} n^pr^{n-1}, for any positive integral choice of p.
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  2. #2
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    Hello, icemanfan!

    I haven't solved this yet, but I have a primitive start on it.
    Maybe someone will be inspired to continue . . .


    Find a way to obtain an explicit formula, in terms of r, such that
    -1 < r < 1, for \sum_{n=1} ^{\infty} n^pr^{n-1}, for any positive integral choice of p.

    For p = 1

    \begin{array}{ccccc}\text{We have:} & S &=& 1 + 2r + 3r^2 + 4r^3 + 5r^4 + \hdots & {\color{blue}[1]} \\<br />
\text{Multiply by }r\!: &rS &=&\qquad r + 2r^2 + 3r^3 + 4r^4 + \hdots & {\color{blue}[2]}\end{array}

    Subtract {\color{blue}[1] - [2]}: . (1-r)S \;=\;1 + r + r^2 + r^3 + r^4 + \hdots

    We have: . (1-r)S \;=\;\frac{1}{1-r}

    Therefore: . \boxed{S \;=\;\frac{1}{(1-r)^2}} ... for p = 1


    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~


    For p = 2

    \begin{array}{ccccc}\text{We have:} & S &=& 1^2 + 2^2r + 3^2r^2 + 4^2r^3 + 5^2r^4 + \hdots & {\color{blue}[1]} \\<br />
\text{Multiply by }r\!: & rS &=&\qquad \;r + 2^2r + 3^2r^3 + 4^2r^4 + \hdots & {\color{blue}[2]}\\\end{array}

    \begin{array}{ccccc}\text{Subtract:} & (1-r)S &=&  1 + 3r + 5r^2 + 7r^3 + 9r^4 + \hdots \\<br />
\text{Multiply by }r\!: & r(1-r)S &=&\qquad r + 3r^2 + 5r^3 + 7r^4 + \hdots  \end{array}

    Subtract: . (1-r)^2S \;=\;1 + 2r + 2r^2 + 2r^3 + 2r^4 + \hdots

    . . (1-r)^2S \;=\;1 + 2r(1 + r + r^2 + r^3 + \hdots) \;=\;1 + 2r\cdot\frac{1}{1-r}

    Therefore: . (1-r)^2S \;=\;\frac{1+r}{1-r} \quad\Rightarrow\quad \boxed{S \;=\;\frac{1+r}{(1-r)^3}} ... for p=2


    . . . Anyone getting excited yet?

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  3. #3
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    Note that (by usual properties of power series), for |r|<1:
    \sum_{n=0}^\infty r^n=\frac{1}{1-r},
    \sum_{n=1}^\infty nr^{n-1}=\frac{d}{dr}\left(\sum_{n=0}^\infty r^n\right)=\frac{d}{dr}\left(\frac{1}{1-r}\right) and, in general,
    \sum_{n=p}^\infty n(n-1)\cdots (n-p+1)r^{n-p}=\frac{d^p}{dr^p}\left(\frac{1}{1-r}\right)=\frac{p!}{(1-r)^{p+1}}.

    ...so that a general method to get a simple expression for \sum_{n=1}^\infty n^p r^{n-1} consists in writing n^p as a linear combination of 1, n, n(n-1),\ldots, n(n-1)\cdots(n-p+1).

    For instance, n^2=n+n(n-1), n^3=n+3n(n-1)+n(n-1)(n-2). And for each of the terms, \sum_{n=1}^\infty n(n-1)\cdots (n-p+1) r^{n-1}=r^{p-1} \sum_{n=p}^\infty n(n-1)\cdots (n-p+1) r^{n-p}=r^{p-1}\frac{p!}{(1-r)^{p+1}}.

    There is no "extremely simple" formula valid for all p, since you need the coefficients in the decomposition n^p=a_1 n +a_2 n(n-1)+\cdots a_p n(n-1)\cdots (n-p+1). These coefficients are called Stirling numbers of the second kind, denoted by \left\{\begin{matrix} p \\ k \end{matrix}\right\}.

    One has n^p=\sum_{k=0}^p\left\{\begin{matrix} p \\ k \end{matrix}\right\} n(n-1)\cdots (n-k+1). A closed formula would then be (up to possible computation errors):
    \sum_{n=1}^\infty n^p r^{n-1}=\sum_{k=0}^p\left\{\begin{matrix} p \\ k \end{matrix}\right\} k!\frac{r^{k-1}}{(1-r)^{k+1}}=\frac{1}{r^2}\sum_{k=0}^p\left\{\begin{m  atrix} p \\ k \end{matrix}\right\} k!\left(\frac{r}{1-r}\right)^{k+1}.
    Last edited by Laurent; October 5th 2008 at 04:27 AM.
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