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?

  • 2
    $\begingroup$ It has a removable singularity at $z=0$. $\endgroup$ – Fly by Night Dec 10 '16 at 14:26
  • $\begingroup$ It is the same as $\frac{\sin z}{z}$ or $\frac{z}{z}$, both well-defined and continuous and analytic at $z=0$ (by $f(0) = \lim_{z \to 0} f(z)$) $\endgroup$ – reuns Dec 10 '16 at 14:36

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, $L < 1$. 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


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.