# Using Hadamard's Formula to show that the radius of convergence of $\sum z^{n!}$ is $1$

Background: Recall that Hadarmard's formula for the radius of convergence of a complex power series $\sum a_n z^n$ is as follows:

$$R = \frac{1}{\underset{n \rightarrow \infty}{limsup} \left| a_n \right|^{1/n}}$$

Goal: Show that the series

$$\sum z^{n!}$$

has radius of convergence equal to $1$ via Hadamard's formula.

Attempt:

1. In order to obtain the limsup in the denominator above, we have to convert the power series $\sum z^{n!}$ into form $\sum a_n z^n$ so that we can check that the expression

$$\underset{n \rightarrow \infty}{limsup} \left|a_n \right|^{1/n}$$

is equal to $1$ as desired.

2. Now consider that

$$z^{n!} = \left(z^{(n!-n)}\right) z^n$$

3. Then if $|z| < 1$, we have that

$$\underset{n \rightarrow \infty}{limsup} \left|z^{(n!-n)} \right|^{1/n} = 1$$

4. If $|z| = 1$, we have that

$$\underset{n \rightarrow \infty}{limsup} \left|z^{(n!-n)} \right|^{1/n} = 1$$

5. If $|z| > 1$, we have that

$$\underset{n \rightarrow \infty}{limsup} \left|z^{(n!-n)} \right|^{1/n} = \infty$$

Given the massive dependence of the $a_n$ on $z$, I'm not sure how to proceed.

• There's no need. If $|z|<1$, it's dominated by a convergent series. If $|z|\ge 1$, the terms don't converge to $0$. Mar 4 '14 at 17:49

If you want to write this as a series, write this as a series! That is, as $\sum\limits_na_nz^n$ where the sequence $(a_n)$ is independent of $z$.

Here $a_n=1$ when $n$ is in $K=\{k!\,;\,k\geqslant0\}$ and $a_n=0$ otherwise, hence $|a_n|^{1/n}$ is always $0$ or $1$ and is $1$ infinitely often (since $K$ is infinite). In particular, $$\limsup\limits_{n\to\infty}|a_n|^{1/n}=1.$$

• And there are much simpler and shorter approaches than Hadamard's formula (a vastly overrated tool, based on the questions on this site).
– Did
Mar 4 '14 at 17:54
• I edited something in your answer, I hope you don't mind. It was a broken code. Mar 4 '14 at 17:57
• @PedroTamaroff Thanks for the edit.
– Did
Mar 4 '14 at 17:58

Hint: All coefficients have $a_n\in\{0,1\}$. No matter how big $n$ is there is a later term with $a_n=1$. Therefore, $$\limsup_{n\to\infty}a_n=1$$

No, the coefficients of your series are $$a_m=\begin{cases}1&\text { if }m=n!\\0&\text{ else }\end{cases}$$

Can you try again with this?