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Thread: 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 $\displaystyle \sum_{n=1} ^{\infty} n^pr^{n-1}$, 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 $\displaystyle r$, such that
    $\displaystyle -1 < r < 1$, for $\displaystyle \sum_{n=1} ^{\infty} n^pr^{n-1}$, for any positive integral choice of $\displaystyle p.$

    For $\displaystyle p = 1$

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

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

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

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


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


    For $\displaystyle p = 2$

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

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

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

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

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


    . . . Anyone getting excited yet?

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

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

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

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

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