Shouldn't we ensure |z|<1 before proceeding further in this question?

Question

Consider, for example, the answer to this question. Here, while converting the series from the denominator using the expansion of $\dfrac 1{1-y}$, why is it not ensured that $|y|< 1$. Here by $y$ I am referring to $(\dfrac{-z^{12}}{12}+\dfrac{z^4}{360}- ...)$. I have seen such conversions from denominator to a series using $1/(1-y)$ often and always seen people not ensuring that $|y|< 1$. Am I missing out something obvious?

Answer 1

All you need is to choose $z$ so that $2|z|^{2} (\frac 1 {4!}+\frac 1 {6!}+\frac 1 {8!}+...) <1$ which is possible since the series $\frac 1 {4!}+\frac 1 {6!}+\frac 1 {8!}+...$ is convergent.

In the second link you only need convergence of $\sum \frac {4^{n}} {(2n)!}$