If the radius of convergence of a function $f(z)$ is $\infty$, is it always true that $f(z)$ is entire?
I always thought that was obvious since any analytic function can be represented with a Taylor series and the radius of convergence being infinite should mean that it is entire. But I've come across this function that is a bit tricky:
$$ f(z) = \frac{e^{z^2}-1}{z^2} $$ $$ \frac{e^{z^2}-1}{z^2}=1+\frac{1}{2!}z^2+\frac{1}{3!}z^4+\frac{1}{4!}z^6+\cdots =\sum_{n=0}^{\infty} \frac{z^{2n}}{(n+1)!} $$
The ratio test tells me that $f(z)$ has the radius of convergence infinity.
However, $f(z)$ doesn't seem to be defined when $z=0$, although according to the series representation $f(0)=1$.
How can this be explained?
A power series has a radius of convergence, not a function.
In your case, $f(z) = \frac{e^{z^2}-1}{z^2}$ is defined on $\Bbb C \setminus \{ 0 \}$, and $$ f(z) = \sum_{n=0}^{\infty} \frac{z^{2n}}{(n+1)!} \quad \text{for all } z \in \Bbb C \setminus \{ 0 \} $$ The radius of convergence of the power series on the right-hand side is $\infty$, therefore
$$ g(z) := \sum_{n=0}^{\infty} \frac{z^{2n}}{(n+1)!} \quad (z \in \Bbb C) $$ is an entire function.
So $f$ is the restriction of an entire function to $\Bbb C \setminus \{ 0 \}$. In other words, $f$ has a removable singularity at $z=0$. This would also follow directly from the existence of the limit $$ \lim_{z \to 0} f(z) = 1 $$ due to Riemann's theorem.
You can extend the definition of $f$ to $\Bbb C$ by setting $f(0) := 1$. Strictly speaking, this is a different function (because it is defined on a different domain). But loosely speaking, the extended function is sometimes called $f$ again, and this would be an entire function.