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?