Here is a property about limits that we were taught in school:
Theorem:
If the limits of functions $f(x),g(x)$ exist at $x_0 \in \mathbb{R}$ then:
$$\lim_{x \to x_0} (f(x) + g(x)) = \lim_{x \to x_0} f(x) + \lim_{x \to x_0} g(x) $$
This can be extended for $x_0 = \infty \blacksquare$
Now I will describe a violation of this property, that is observed in the computation of Z-transform(ZT) of a 2-sided sequence $x[n], n \in \mathbb{Z}$, where $x[n] \neq 0$ for some $n_1 < 0$ and for some $n_2 > 0$.
$$\sum_{n =-\infty}^{\infty} x[n] z^{-n} = \sum_{n =-\infty}^{-1} x[n] z^{-n} + \sum_{n = 0}^{\infty} x[n] z^{-n} $$ We know how to compute each of the 2 new sums separately. If the first sum converges for $|z| < r_{+}$, and the second sum converges for $|z| > r_{-}$. The Z-Transform exists at least for all $z$ such that $r_{-} < |z| < r_{+}$ given that $r_{-} < r_{+}$. Given the above assumptions, we have found a subset of the Region of Convergence(ROC) of the ZT.
But if $r_{-} \geq r_{+}$ textbooks mention that the Z-Transform does not exist! However, the Theorem stated above implies that when the infinite sums do not exist simultaneously, they cannot be broken apart. Thus we cannot claim non-existence of ZT since the initial sum may converge.
Example: I tried computing the ZT-transform of the following sequence, assuming that $|a| \geq |b|$. $$x[n] = \left\{ \begin{array}{ll} a^n & n \geq 0\\ -b^n & n < 0 \end{array} \right.$$ The result was that for some discrete antipodal points $z_1,z_2$ in the complex plane $X(z)$ can be computed. Of course in this case 2 numbers $X(z_1), X(z_2)$ cannot describe a sequence of infinite terms...
Hence, it might be the case that for many/common sequences we can only compute their ZT @ discrete finitely many points. And thus ZT doesn't carry any useful information about the original sequence. Thus the ZT is practically non-existent. But such assumptions should be proved rigorously.
1. So is there some theory behind this?
2. Can you point out any mistakes I made in the thought process?
Thanks in advance!
EDIT: Here the correspondence to the limit property is clarified. $$\sum_{n = -\infty}^{\infty} x[n]z^{-n} = \lim_{N \to \infty} \sum_{n = -N}^{N}x[n]z^{-n} = \lim_{N \to \infty} \Biggl \{ \sum_{n = -N}^{-1}x[n]z^{-n} + \sum_{n = 0}^{N}x[n]z^{-n} \biggr \}$$
Unlike one-sided series, convergence of two-sided series of the form $\sum_{n=-\infty}^\infty a_n$ is only unambiguously defined if $\sum_{n=-\infty}^\infty |a_n|<\infty$. In this case, if either $\sum_{n=0}^\infty |x[n]z^{-n}|=\infty$ or $\sum_{n=-\infty}^{-1} |x[n]z^{-n}|=\infty$, then $\sum_{n=-\infty}^\infty |x[n]z^{-n}|=\infty$. This will happen when $r_->r_+$. So convergence of $\sum_{n=-\infty}^\infty x[n]z^{-n}$ without separate convergence of $\sum_{n=0}^\infty x[n]z^{-n}$ and $\sum_{n=-\infty}^{-1} x[n]z^{-n}$ isn't possible.
There is a temptation to believe that convergence of $\lim_N \sum_{n=-N}^N a_n$ should be how convergence of $\sum_{n=-\infty}^\infty a_n$ is defined, but it isn't. If $\sum_{n=-\infty}^\infty a_n$ is unambiguously defined, it should be independent of rate of approach of $\infty$. That is, if we say that $\sum_{n=-\infty}^\infty a_n$ is convergent, it should be the case that for any $M_k,N_k\in \mathbb{N}$ with $\lim_k M_k=\lim_k N_k=\infty$, $\lim_k \sum_{n=-M_k}^{N_k}a_n=\lim_N \sum_{n=-N}^N a_n$.
To see why $\sum_{n=-\infty}^\infty a_n$ can be ambiguous when $\sum_{n=-\infty}^\infty |a_n|=\infty$, let $a_n=1$ if $n>0$, $a_n=-1$ if $n<0$, and $a_0=0$. Then $\sum_{n=-N}^N a_n=0$ for any $N\in \mathbb{N}$, but $\sum_{n=-N}^{N+1}a_n=1$. So if the upper and lower bounds on the partial sums approach $\pm \infty$ in different ways, you can get different results. This isn't an issue when you only have a one-sided sum.