I am working on this problem and struggling to apply the power series.
EDIT: splitting $(z+1)/(z-1)$ into $z/(z-1) + 1/(z-1)$ and then using a table of $z$ transform might work?
EDIT:This is my current solution/attempt \begin{aligned} X(z) &= \frac{z+1}{z-1}\\ &= 1 + \frac {1}{z} * \frac{1}{1+1/z} \end{aligned} Then using the power series for the 2nd part of the eqn: $$ = 1 + 1/z\left(\frac {1}{1-z^{-1}}\right) $$
$n0$: $1$
$n1$: $1$
$n2$: $2$
$n3$: $1.5$
$n4$: $1.3$
This is just my current attempt, I apologise if i have used mathjax incorrectly, its my first attempt!
$$\frac{z+1}{z-1}=\frac{1+\frac{1}{z} }{1-\frac{1}{z}}=\left(1+\frac{1}{z}\right)\sum_{n=0}^\infty \frac{1}{z^n} =\sum_{n=0}^\infty \frac{1}{z^n}+\sum_{n=0}^\infty \frac{1}{z^{n+1}}$$ $$\frac{z+1}{z-1}=1+\sum_{n=1}^\infty \frac{2}{z^{n}}$$ $$\mathcal{Z}^{-1}\left(\frac{z+1}{z-1}\right)_{(n)}=\begin{cases}1 \qquad \text{if}\quad n=0 \\2 \qquad\text{if}\quad n \geq 1 \end{cases} $$ $\mathcal{Z}^{-1}$ denotes the inverse Z-transform.