Is a signal summable when given a z-transform?

Question

The output of a system is given as z-transform: $$ Y(z)=\frac{1+z^{-2}}{(1+\frac{1}{4}z^{-1})(1-\frac{1}{2}z^{-1})} $$ I want to know if the signal in the time domain is summable, meaning that the following is true: $$ \sum_{n=-\infty}^{\infty}|y[n]|<B<\infty $$ Is it enough to say: Well, there is already a $Y(z)$ given, so the infinite sum for calculating a z-transform obviously converges, which means $y[n]$ is summable? I am not sure about this, because when calculationg a z-transform the z also plays a major role if the sum converges or not: $$ Y(z)=\sum_{n=-\infty}^{\infty}y[n]z^{-n} $$ How do I do this correctly?