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.


  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.

  • 2
    $\begingroup$ There's no need. If $|z|<1$, it's dominated by a convergent series. If $|z|\ge 1$, the terms don't converge to $0$. $\endgroup$ – David Mitra 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, $$\limsup\limits_{n\to\infty}|a_n|^{1/n}=1.$$

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


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