Z transform convergence
Nop. For an infinite sequence, the convergence of Z transform depends on the value of $z$ also. So as far as $n\rightarrow\pm\infty$, equation (1) will not be satisfied for all values of $z$.

Consider a simple case, $x(n) = u(n)$
$$X(z) = \sum_{n=0}^{\infty}z^{-n}$$
$$X(z) = 1 + z^{-1}+z^{-2}+z^{-3}+ \cdots$$
for $z=0.5$,
$$X(z) = 1 + 2+4+8+ \cdots = \infty$$
for this sequence to converge, $|z|&gt;1$.