# 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?

• It has a removable singularity at $z=0$. – Fly by Night Dec 10 '16 at 14:26
• It is the same as $\frac{\sin z}{z}$ or $\frac{z}{z}$, both well-defined and continuous and analytic at $z=0$ (by $f(0) = \lim_{z \to 0} f(z)$) – reuns Dec 10 '16 at 14:36

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, $L < 1$. You got $0$ so lets check
$$0 < 1$$ $0$ will always be less than $1$, so you can say that is absolute convergence, that result in convergence and is for all $Z$ values