# Pseudo-Geometric Series

• Oct 4th 2008, 05:59 PM
icemanfan
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.
• Oct 4th 2008, 07:03 PM
Soroban
Hello, icemanfan!

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

Quote:

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 $p = 1$

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

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 $p = 2$

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

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

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?

• Oct 5th 2008, 03:53 AM
Laurent
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):
$\sum_{n=1}^\infty n^p r^{n-1}=\sum_{k=0}^p\left\{\begin{matrix} p \\ k \end{matrix}\right\} k!\frac{r^{k-1}}{(1-r)^{k+1}}=\frac{1}{r^2}\sum_{k=0}^p\left\{\begin{m atrix} p \\ k \end{matrix}\right\} k!\left(\frac{r}{1-r}\right)^{k+1}$.