Z transform convergence From the definition I know:

$$|\sum_{n=-\infty}^{+\infty}x[n] \cdot z^{-n}| &lt; \infty\space\space\space(1)$$

Is there any sequence of x[n] which can not be written as 

$$x[n] = y[n] \cdot u[an-b]\space,\space\space\space a,b \in \mathbb{Z}$$
(where the dot means a multiplication and $u[k]$ means the discrete step function. In other words, x[n] is an infinity sequence)

but do satisfy the definition in (1)?
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, $|z|&gt;1$.