What do they mean by radius of convergence?

Question

I am looking at a couple different page on the definition of Radius of Convergence, specifically for Taylor series. I first learned it as follows: For a power series $$\sum_{k=0}^\infty a_k (z-z_0)^k$$ the radius of convergence is a unique real number $R\in\mathbb R \cup \{0,\infty\}$ where the sum converges when $|z-z_0|<R$ and diverges when $|z-z_0|>R$.

However, I was told a different definition/convention specific to Taylor series: For a function $f:U\to\mathbb C$ and any $z_0\in U$ we say the radius of convergence for the Taylor series centered at $z_0$ is the largest $R$ for which the Taylor series converges to $f$ on $D(z_0;R)$. So it not only needs to converge, it has to converge to $f$.

This two definitions are clearly different, for example, consider the function $g:\mathbb C\setminus\{1\}$ where $g(z)=0$. The Taylor series of $g$ centered at $z_0=0$ is $0$. Using the first definition we know that the radius of convergence of this Taylor series is $\infty$. Using the second definition we see that the radius of convergence is actually $1$, since it does not converge to $g$ at $z=1$.

This is confusing already but upon a bit of searching, it seems like both of these definition contradicts with this fact in the wikipedia article: https://en.wikipedia.org/wiki/Analyticity_of_holomorphic_functions which says: "the radius of convergence is always the distance from the center $a$ to the nearest non-removable singularity;"

Using the same $g$ as above, this fact produces contradictory result with the second definition. Consider $h:\mathbb C\setminus \mathbb R^-\to \mathbb C$ where $h(z)=0$, $h$ has no removable singularity since none of them are isolated. Now the Taylor series of $h$ at $z_0=1$ is $0$ and has radius of convergence $\infty$. However, the distance between $z_0$ and the nearest non-removable singularity is $1$, since $\mathbb R^-$ is a set of non-removable singularities. So $h$ shows that the fact is contradictory with the first definition.

Given those, I have the following questions:

1. Is anything I have stated incorrect?
2. Is it a convention to define radius of convergence differently for Taylor series? (instead of the series converging, it has to converge to the function where the series is defined on)
3. Is the fact listed in wikipedia correct? Is it yet another convention for "radius of convergence"?

Answer 1

Radius of convergence is a property of a power series, not of a function. Your first definition is correct, your second is not. The Wikipedia statement is misleading. What is true is that if the radius of convergence is $R$ (with $0 < R < \infty$), the Taylor series converges on the open disk of radius $R$ centered at $a$ to a function $f$ analytic on that disk, but it is impossible to extend $f$ to be analytic on a disk centered at $a$ with any larger radius; this implies that there is some point $p$ at distance $R$ from $a$ such that there is no way to extend $f$ to be analytic in a neighbourhood of $p$. In that sense, $R$ is the distance to the closest non-removable singularity. However, if you start with a function $g$ that has the same Taylor series, there is no reason to expect $g$ to agree with $f$ on that disk.

[EDIT] The exception is if $g$ is assumed to be analytic everywhere except for isolated singularities. Then the radius of convergence is, in fact, the distance from $a$ to the closest non-removable singularity of $g$. On the other hand if, say, $g$ has a branch cut, there is no reason that branch cut can't wander inside the disk of radius $R$ around $a$.