1
$\begingroup$

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.

$\endgroup$
  • 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
4
$\begingroup$

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, $$\limsup\limits_{n\to\infty}|a_n|^{1/n}=1.$$

$\endgroup$
  • $\begingroup$ 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
2
$\begingroup$

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, $$ \limsup_{n\to\infty}a_n=1 $$

$\endgroup$
2
$\begingroup$

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?

$\endgroup$

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.