Prove that for $z \in\mathbb C\setminus \{1\},\, 1 + z + z^2 +... + z^n = (1-z)^{n+1}/(1 - z)$ for $n = 1,2,3\ldots$

Question

Prove that for $z \in\mathbb C\setminus \{1\}$,

$$\ 1 +z+z^2+...+z^n=\frac{1- z^{n+1}}{1-z}\quad (n = 1,2,3\ldots)$$

This is a complex analysis problem. I have been out of an official math class for a long time and am struggling to even know where to approach a problem like this, more or less where to start. Any help is greatly appreciated.

Multiply both sides by $1-z$ to get an obviously true statement, then just reverse the process for the proof. — saulspatz, Aug 29 '18 at 03:40
Call the sum $s_n$. Now consider $z s_n$. What happens when you minus $z s_n$ from $s_n$? — Mo Pol Bol, Aug 29 '18 at 03:45
Your title is not correct as the $n+1$ power is applied to $(1-z),$ not just to $z$ — Ross Millikan, Aug 29 '18 at 03:53
@RossMillikan That might be the case. I transcribed according to the problem in the homework — Mr. Dole, Aug 29 '18 at 04:00
@Mr.Dole What Ross Millikan pointed out is that your title does not match the body of the question. — dxiv, Aug 29 '18 at 04:13

score 0 · Answer 1 · answered Aug 29 '18 at 03:45

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HINT: What is $$ (1+z+z^2+\ldots+z^n)(1-z)? $$

answered Aug 29 '18 at 03:45

Przemysław Scherwentke

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$$\sum_{k=0}^{n-1}z^k(1-z)$$ – Mr. Dole Aug 29 '18 at 03:56
@Mr.Dole Multiply the first expression by 1, then by $-z$ and reduce. – Przemysław Scherwentke Aug 29 '18 at 03:58
Sorry I have been struggling with formatting. Is this the form you are looking for? – Mr. Dole Aug 29 '18 at 03:59
@Mr.Dole Might be when you can see what remains after reduction. You can guess what should be. :-) – Przemysław Scherwentke Aug 29 '18 at 04:01
Sorry my brain is slow... Am I supposed to pull out a 1-z? I don't know how to reduce further :/ – Mr. Dole Aug 29 '18 at 04:05
@Mr.Dole $(1+z+z^2+\ldots+z^n)-(z+z^2+\ldots+z^n+z^{n+1})$ and the best for beginning is to write it in two lines: $z$ below $z$, $z^2$ below $z^2$ etc. – Przemysław Scherwentke Aug 29 '18 at 04:12
I see now thank you so much! – Mr. Dole Aug 29 '18 at 04:23

score 0 · Answer 2 · answered Aug 29 '18 at 04:39

This is an example of "telescoping", since the center of the series collapses and we get $(1-z)(1+z+z^2+\cdots +z^n)=1-z+z-z^2+z^2+\cdots -z^{n-1}+z^{n-1}-z^n+z^n-z^{n+1}=1-z^{n+1}$. Just divide by $(1-z)$ to complete the proof.

Prove that for $z \in\mathbb C\setminus \{1\},\, 1 + z + z^2 +... + z^n = (1-z)^{n+1}/(1 - z)$ for $n = 1,2,3\ldots$

2 Answers2