summation proving

**TechnicianEngineer** · February 6th 2011, 12:08 AM

how do i prove
$-\displaystyle \sum_{n=-\infty}^{-1} \alpha^{n} z^{-n} =-\displaystyle \sum_{n=0}^{\infty} (\alpha^{-1}z)^{n+1}$

???

**tonio** · February 6th 2011, 01:00 AM

Originally Posted by TechnicianEngineer

how do i prove
$-\displaystyle \sum_{n=-\infty}^{-1} \alpha^{n} z^{-n} =-\displaystyle \sum_{n=0}^{\infty} (\alpha^{-1}z)^{n+1}$

???

$-\displaystyle{\sum_{n=-\infty}^{-1} \alpha^{n} z^{-n}=-\sum_{n=-\infty}^{-1}\left(\alpha^{-n} z^{n}\right)^{-1}=-\sum_{n=-\infty}^{-1}\left( \alpha^{-1} z\right)^{-n}=-\sum_{n=1}^{\infty}\left(\alpha^{-1} z\right)^{n}}$

Tonio

**HallsofIvy** · February 6th 2011, 03:42 AM

Slight variation: Let j= -n. Then $\alpha^n= \alpha^{-j}$ and $z^{-n}= z^j$ . When n= -1, j= 1, as n goes to $-\infty$ , j goes to $\infty$ .

$-\sum_{n= -\infty}^{-1}\alpha^nz^{-n}= -\sum_{j= 1}^\infty\alpha^{-j}z^j= -\sum_{j= 1}^\infty \left(\alpha^{-1}z)^j$ and now just change the index from "j" to "n".

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