1. PF Contest - Win "Conquering the Physics GRE" book! Click Here to Enter
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Converting 1/(1-z) to a sum

  1. Dec 17, 2015 #1
    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?

    (2. Relevant equations)

    (3. The attempt at a solution)
  2. jcsd
  3. Dec 17, 2015 #2


    Staff: Mentor

    ##\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.
  4. Dec 17, 2015 #3
    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
  5. Dec 22, 2015 #4


    User Avatar
    Science Advisor

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

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

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

    Essentially, the "[itex](-1)^n[/itex]" comes from the fact that this is [itex]\frac{1}{1+ (z- 1)}[/itex] rather than [itex]\frac{1}{1- (z-1)}[/itex] which would have power series [itex]\sum_{n=0}^\infty (z- 1)^n[/itex].
    Last edited by a moderator: Dec 22, 2015
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?
Draft saved Draft deleted