Prove complex equation

Question

Prove the following $$\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$$

for any integer n greater than 0. DO NOT USE ....

I believe that I can use mathematical induction. Any help would be greatly appreciated.

Answer 1

And here's a proof by induction.

It is true for $n=0$ when it is $\frac1{1-z} = \frac1{1-z} $.

Suppose it is true for $n$. We want to show that it is true for $n+1$, or that $\frac{1}{z-1}*\frac{1}{z^{n+1}} = \dfrac{1}{z-1} - \sum_{k=1}^{n+1}\frac{1}{z^k} $.

Using the induction hypothesis that $\frac{1}{z-1}*\frac{1}{z^{n}} = \dfrac{1}{z-1} - \sum_{k=1}^{n}\frac{1}{z^k} $

$\begin{array}\\ \dfrac{1}{z-1} - \sum_{k=1}^{n+1}\frac{1}{z^k} &=\dfrac{1}{z-1} - \left(\sum_{k=1}^{n}\frac{1}{z^k}+\frac1{z^{n+1}}\right)\\ &=\left(\dfrac{1}{z-1} - \sum_{k=1}^{n}\frac{1}{z^k}\right)-\frac1{z^{n+1}}\\ &=\left(\frac{1}{z-1}*\frac{1}{z^{n}}\right)-\frac1{z^{n+1}} \text{ (induction hypothesis used here)} \\ &=\frac{z-(z-1)}{(z-1)z^{n+1}}\\ &=\frac{1}{(z-1)z^{n+1}}\\ \end{array} $

and the result is true for $n+1$.

Answer 2

You want to show that $\frac{1}{z-1}*\frac{1}{z^n}= \dfrac{1}{z-1} - \sum_{k=1}^n\frac{1}{z^k}$

Multiply by $z-1$ and it becomes $\frac{1}{z^n} = 1 - (z-1)\sum_{k=1}^n\frac{1}{z^k} $ or $1-\frac{1}{z^n} = (z-1)\sum_{k=1}^n\frac{1}{z^k} $.

But

$\begin{array}\\ (z-1)\sum_{k=1}^n\frac{1}{z^k} &=z\sum_{k=1}^n\frac{1}{z^k}-\sum_{k=1}^n\frac{1}{z^k}\\ &=\sum_{k=0}^{n-1}\frac{1}{z^k}-\sum_{k=1}^n\frac{1}{z^k}\\ &=1-\frac1{z^n}\\ \end{array} $

QED