Radius of convergence and analyticity 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.
You need to check the convergence with ratio test
$$\require{cancel}L=\lim\limits_{n \to \infty} \left|{\frac{f(n+1)}{f(n)}}\right|$$
That results in
$$L=\lim\limits_{n \to \infty} \left|{{z^{{2n}+2}\over (n+2)!}\over {z^{2n}\over {(n+1)!}}}\right|$$
$$L=\lim\limits_{n \to \infty} \left|{{z^{2+\cancel{2n}}\over (n+2)\cdot\cancel{(n+1)!}}\over {z^\cancel{2n}\over \cancel{(n+1)!}}}\right|$$
$$L=\lim\limits_{n \to \infty} \left|{{z^{2}\over (n+2)}\over {1\over 1}}\right|$$
$$L=\lim\limits_{n \to \infty} \left|{{z^{2}\over (n+2)}\over {1\over 1}}\right|$$
$$L=\lim\limits_{n \to \infty} \left|{z^2\over (n+2)}\right|$$
$$L=|z^2|\lim\limits_{n \to \infty} \left|\cancelto{0}{1\over (n+2)}\right|$$
$$L=0$$
So for convergence, $L &lt; 1$. You got $0$ so lets check

$$0 &lt; 1$$
$0$ will always be less than $1$, so you can say that is absolute convergence, that result in convergence and is for all $Z$ values