Why does (-1)^n appear in the power series for 1/(1+(z-1))?

[Jon.G]

Homework Statement

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. Homework Equations )(3. The Attempt at a Solution )

 

[Mark44]

Homework Statement

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. Homework Equations )(3. The Attempt at a Solution )

$\frac 1 {1 + u} = 1 - u + u^2 - u^3 +- \dots (-1)^n u^n + \dots$
BTW, the expression in your thread title is different from what you actually asked about.

 

[Jon.G]

BTW, the expression in your thread title is different from what you actually asked about.

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

 

[HallsofIvy]

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.