How to calculate Z-transform from a series

Question

If we have the sequence $\{1,1,\dots\}$, how would I go about calculating the Z-transform? Such that we find $Z\{1,1 \dots\}(z)$.

Answer 1

$1+1/z+1/z^2+\cdots= 1/(1-1/z)=z/(z-1)$

Answer 2

$$Z\{1,1 \dots\}(z)=\dfrac {z}{z-\color{red}{1}z}=\dfrac {1}{1-\color{red}{1}z^{-1}}=\sum_{n=0}^{\infty}\color{red}{1^n}z^{-n}$$ $$Z\{1,2,4 \dots\}(z)=Z\{2^0,2^1,2^2 \dots\}(z)$$ $$Z\{2^0,2^1,2^2 \dots\}(z)=\dfrac {z}{z-\color{red}{2}z}=\dfrac {1}{1-\color{red}{2}z^{-1}}=\sum_{n=0}^{\infty}\color{red}{2^n}z^{-n}$$ $$Z\{2,2,2 \dots\}(z)=\dfrac {2z}{z-\color{red}{1}z}=\dfrac {2}{1-\color{red}{1}z^{-1}}=2\sum_{n=0}^{\infty}\color{red}{1^n}z^{-n}$$