# Z Transform - ROC values

I was reading a DSP book where they had a problem on Z Transform.

Determine the z-transform and ROC for signal $\displaystyle x(n) = a^nu(n)$?

The solution states that $$\displaystyle X(z) = \displaystyle\sum_{n=0}^\infty (az{^{-1}})^n$$ So $$\displaystyle X(z) = \frac{ \displaystyle {z} }{z-a}$$ Hence ROC $|z| > a$

Which I understand that for all values of z > a the series $\frac{ \displaystyle {z} }{z-a}$ has finite values.

But for $|z| < a$ their exists a value : $z = 5$ and $a = 10$ the series has a finite value of -1 and so on. So why -1 is not taken as a set for which ROC exists - or rather my understanding on ROC is wrong?

The fact that evaluating the function \begin{align} \frac{z}{z-a} \end{align} yields a finite value for some given values of $z$ and $a$, does not mean that the original series (from $\mathcal{Z}$-transform definition) \begin{align} X(z) = \sum_{n=0}^\infty(az^{-1})^n \end{align} converges at all for those same values of $z$ and $a$. Before we determine the region of convergence, the only thing we can say is that if the series converges, then it will converge to $\frac{z}{z-a}$.
In the case $z=5$ and $a=10$, looking at the sequence \begin{align} \left\{\sum_{n=0}^N (az^{-1})^n \right\} \end{align} as $N$ increases yields: $$\{1, 3, 7, 15, 31, 63, ...\}$$ which as you may notice does not converge to -1, and in fact diverges as $N\rightarrow\infty$. The region of convergence tells you where this sequence converges (and in this case it is $|z|>|a|$).