# Converting 1/(1-z) to a sum

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1. Dec 17, 2015

### Jon.G

1. The problem statement, all variables and given/known data
So I'm checking my solutions to past question and there's one bit that throws me.
1/(1+(z-1)) = Σ(-1)n(z-1)n (for 0<|z-1|<1)
I don't know where the (-1)n factor came from. Is it just something that always happens that I didn't know about / forgot about, or is there some other reason that I'm missing?
Thanks

(2. Relevant equations)

(3. The attempt at a solution)

2. Dec 17, 2015

### Staff: Mentor

$\frac 1 {1 + u} = 1 - u + u^2 - u^3 +- \dots (-1)^n u^n + \dots$

3. Dec 17, 2015

### Jon.G

That just made me realize DX
The expression 1/(1+(z-1)) is the same as 1/(1-(-(z-1)))
and the general rule is: 1/(1-x) = Σxn
That double minus is where the -1 factor came from.
I think I understand it, now sure if I explained why I understand it well enough (so many brackets). Just missed out on the sign :D

4. Dec 22, 2015

### HallsofIvy

Staff Emeritus
It helps to know that the sum of a "geometric series", $\sum_{n= 0}^\infty ar^n$ is $\frac{a}{1- r}$. That can be shown by looking a the finite series, $S= \sum_{n=0}^N ar^n= a+ ar+ ar^2+ \cdot\cdot\cdot+ ar^N$. We can write that as $S= a+ \left(ar+ ar^2+ \cdot\cdot\cdot+ ar^N\right)= a+ r\left(a+ ar+ \cdot\cdot\cdot+ ar^{N-1}\right)$. The quantity in the parentheses on the right is almost "S" again! We can make it so by adding $ar^N$ inside the parentheses and subtracting $ar^{n+1}$ outside the parentheses: $S= a+ r\left(a+ ar+ \cdot\cdot\cdot+ ar^{N-1}\right)+ ar^N)- ar^{N+1}= a+ r\left(S\right)- ar^{N+1}$. Subtracting $rS$ from both sides, $S- rS= (1- r)S= a- ar^{N+1}= a(1- r^{N+1})$. Finally divide both sides by 1- r to get $S= \frac{a(1- r^{N+1})}{1- r}$.

As long as |r|< 1, we can take the limit as N goes to infinity. Then $r^{N+1}$ goes to 0 and $\sum_{n= 0}^\infty ar^n= \frac{1}{1- r}$.

We can think of $\frac{1}{1+ (z- 1)}$ as $\frac{1}{1- (1- z)}$ and write that as a geometric series with a= 1 and r= 1- z: $\sum_{n=0}^\infty (1- z)^n= \sum_{n=0}^\infty ((-1)(z- 1))^n= \sum_{n=0}^\infty (-1)^n (z- 1)^n$.

Essentially, the "$(-1)^n$" comes from the fact that this is $\frac{1}{1+ (z- 1)}$ rather than $\frac{1}{1- (z-1)}$ which would have power series $\sum_{n=0}^\infty (z- 1)^n$.

Last edited: Dec 22, 2015