I am having trouble demonstrating the $Z$ transform of $a^{n-1}u(n-1)$ is $\frac{1}{(z-a)}$, as it says in this table.
I try using the definition of the z transform, but it comes out different than what the table says:
$$\sum_{k=-\infty}^{\infty}a^{k-1}u(k-1)z^{-k}=\sum_{k=1}^{\infty}a^{k-1}z^{-k}=\frac{1}{a}\sum_{k=1}^{\infty}(\frac{a}{z})^k$$
This is a geometric series, so the result should be:
$$\frac{1}{a}\frac{1}{1-\frac{a}{z}}=\frac{z}{a(z-a)}$$
Why is my answer coming out different than what it says in the table?
The geometric series $\sum_{k\geq k_0} r^k$ converges if and only if $|r|<1$. You have a confusion regarding its actual sum, which depends where the summation starts. In particular $$ \sum_{k\geq 1}r^k=\frac{r}{1-r}\qquad\mbox{and}\qquad\sum_{k\geq 0}r^k=\frac{1}{1-r}. $$ More generally, $$ \sum_{k\geq k_0}r^k=\frac{r^{k_0}}{1-r}. $$ To see where this all comes from, you just need to observe that the partial sums satisfy, for $K\geq k_0$: $$ \sum_{k=k_0}^Kr^k=r^{k_0}(1+r+\ldots+r^{K-k_0})=r^{k_0}\left(\frac{1-r^{K-k_0+1}}{1-r}\right)=\frac{r^{k_0}-r^{K+1}}{1-r}. $$
In your case, you do get $$ \frac{1}{a}\sum_{k\geq 1}\left(\frac{a}{z}\right)^k=\frac{1}{a}\cdot \frac{\frac{a}{z}}{1-\frac{a}{z}}=\frac{1}{z-a}. $$