Problem 2. Determine the z-transform for the unit step sequence $\left\{u_{k}\right\}=\{1,1,1,1, \ldots\}=\{1\}$
The z-transform of $\left\{u_{k}\right\}$ is given by:
$$
\begin{aligned}
Z\left\{u_{k}\right\} &=F(z)=\sum_{k=0}^{\infty} \frac{u_{k}}{z^{k}}=\sum_{k=0}^{\infty} \frac{1}{z^{k}} \\
&=\frac{1}{z^{0}}+\frac{1}{z^{1}}+\frac{1}{z^{2}}+\frac{1}{z^{3}}+\frac{1}{z^{4}}+\ldots
\end{aligned}
$$
i.e. $\quad \mathbf{Z}\left\{u_{k}\right\}=1+\frac{1}{z}+\frac{1}{z^{2}}+\frac{1}{z^{3}}+\frac{1}{z^{4}}+\ldots$
Using the binomial theorem for $(1+x)^{n}$, the series expansion of $\frac{1}{1-x}$ may be determined:
$$
\begin{aligned}
\frac{1}{1-x}=&(1-x)^{-1} \\
=& 1+(-1)(-x)+\frac{(-1)(-2)}{2 !}(-x)^{2} \\
&+\frac{(-1)(-2)(-3)}{3 !}(-x)^{3}+\ldots \\
&=1+x+x^{2}+x^{3}+\ldots \text { valid for }|x|<1
\end{aligned}
$$
Comparing equations (1) and (2) gives: $\mathrm{F}(\mathrm{z})=\frac{1}{1-\frac{1}{z}}$ provided $\left|\frac{1}{z}\right|<1$
$\frac{1}{1-\frac{1}{z}}=\frac{1}{\frac{z-1}{z}}=\frac{z}{z-1}$ hence, $z\left\{u_{k}\right\}=\frac{z}{z-1}$ provided $|z|>1$