How $|\phi| < 1$ and $|z| \leq 1$ is important to prove the convergence of the following power series

Question

Let $f(z) = \sum_{n=0}^\infty (\phi z)^n$.

How can I show that the conditions $|\phi| < 1$ and $|z| \leq 1$ are necessary to prove $f(z)$ converges to $$\frac{1}{1 - \phi z}$$

where $z$ is a complex number.

My approach: From the ratio-test $f(z)$ to converge we need \begin{align*} lim_{n \rightarrow \infty }\left|\frac{(\phi z) ^{n+1}}{(\phi z) ^{n}}\right| = |\phi||z| < 1 \end{align*}

Then if $|\phi| < 1$ then how does it imply $|z| \leq 1$?

Edited:

Another lecture note which uses the same idea:

Answer 1

The series can converge without the conditions $|\phi | <1$ and $|z| \leq 1$. For example take $\phi =\frac 1 4$ and $z=2$.

Answer 2

I suppose that $ \phi \ne 0.$ Then the radius of convergence of the power series is $R= \frac{1}{|\phi|}.$

If, for example, $ |\phi|<1$, then $R>1$, hence the power series converges for $|z| \le 1.$

Answer 3

The idea is that for the series to converge for all $z$ in the unit disk, boundary included, you will need $|ϕ|<1$.

For operators it is the same, for the Neumann series for $(ϕB)$ to converge for all $B$ with $\|B\|\le 1$, you need $ϕ|<1$.

In both cases the set of convergence will be actually larger. This is necessary, as the convergence condition gives an open set, but the claim is about a closed set contained in it. Thus the radius of the closed set has to be smaller than the radius of the containing open set.