What is the Z-transform of the sequence $-b^nu(1-n)$ and what is its ROC? What is the Z-transform of the sequence $-b^nu(n-1)$ and what is its ROC?
Note: $u(n)$ is the unit step sequence.
My reasoning and possible solution is the following:
$$X(Z)= \sum_{k = - \infty}^{-1}(-b)^kz^{-k}=\sum_{k = 1}^{\infty}(-b)^{-k}z^k = \sum_{k = 1}^{\infty}\left ( \frac{-z}{b} \right )^k = -1 + \sum_{k = 0}^{\infty}\left ( \frac{-z}{b} \right )^k =-1 + \frac{1}{1+\frac{z}{b}} = -1 + \frac{b}{z + b} = \frac{-z}{z+b}$$
Assuming that $\left | z \right | < \left | b \right |$ which would be the ROC.

$$x(n) = -(b^n) u(1-n)$$
$$X(z) = \sum_{n=-\infty}^{\infty} x(n) z^{-n}$$
$$X(z) = -\sum_{n=-\infty}^{1}\frac{b^n}{z^n} $$
$$X(z) = - ( \frac{b}{z}+1+\frac{z}{b}+\frac{z^2}{b^2}+......)$$
Now apply the formula for sum of infinte GP
$$X(z) = -\frac{b}{b-z}$$
$$X(z) = \frac{b}{z-b}$$ 
$$|\frac{z}{b}|<1$$
$$ROC :|z|<|b|$$
