Convergence of $\sum_{n=0}^\infty z^{2^n}$

Question

Let formally $f(z) := \sum_{n=0}^\infty z^{2^n}$. What is the raduis of convergence of this series ?

Do you know the Cauchy-Hadamard formula for the radius of convergence? — Daniel Fischer, Jun 28 '13 at 11:28
yes but here I must have something like $\sum_n a_n z^n$ s.t. then $1/R = \limsup |a_n|^{1/n}$ as $n \rightarrow \infty$. — , Jun 28 '13 at 11:30
Well, you can write $\sum_{n=0}^\infty z^{2^n}$ in the form $\sum_{k = 0}^\infty a_k z^k$. You just need to determine the $a_k$. — Daniel Fischer, Jun 28 '13 at 11:31
Different but related question about lacunary series I thought was fun to mention. — Raskolnikov, Jun 28 '13 at 11:53

score 4 · Accepted Answer · answered Jun 28 '13 at 11:50

4

There is no need to think of how to apply any special test or theorem to this. Just think about how the terms grow, and how it compares to the most basic series you've seen before e.g. geometric series. If $|z|<1$ then the size of the terms drops off very quickly. Prove the series converges absolutely (and hence converges) if $|z|<1.$ Now see what happens where $z=1$ and this will determine the radius of convergence.

answered Jun 28 '13 at 11:50

Ragib Zaman

36,065

Well said. It is not even necessary to rewrite this as a series $\sum\limits_na_nz^n$. (A similarly lacunary series was posted a while ago on the site.) – Did Jun 28 '13 at 11:54

score 1 · Answer 2 · edited Jun 28 '13 at 12:07

1

Your series is

$$\sum_{k=0}^\infty a_kz^k\;,\;\;a_k=\begin{cases}0&,\;\;k\;\;\text{is not a power of}\;\;2\\{}\\1&,\;\;\text{otherwise}\end{cases}$$

Now, you can try the usual ways to find out the convergence radius, say

$$\sqrt[k]{a_k}=\begin{cases}0&,\;\;k\;\;\text{is not a power of}\;\;2\\{}\\1&,\;\;\text{otherwise}\end{cases}$$

And thus the $\,\text{sup lim}\;$ of the above is....

edited Jun 28 '13 at 12:07

Cameron Buie

105,149

answered Jun 28 '13 at 11:33

DonAntonio

214,715

2

Is it not $a_k = 1$ if $k$ is a power of $2$ ? – Jun 28 '13 at 11:35
This would give $\limsup_{n \rightarrow \infty} |a_n|^{1/n} = \inf_{m \geq 0} \sup_{n \geq m} |a_n|^{1/n} = \inf_{n \geq 0} 1 = 1$ I guess. – Jun 28 '13 at 11:40
1

I suck today way more than I usually do...thanks all. – DonAntonio Jun 28 '13 at 11:47
We have $\sum_n z^{2^n} = z+z^2+z^4+z^8 + \cdots$ Why is this equal to $\sum_n a_kz^k$ ? – Jun 28 '13 at 11:49
1

Made a few fixes for you. (Hope you don't mind.) I wasn't sure what "Bow" was supposed to be, though. – Cameron Buie Jun 28 '13 at 11:50
1

Ok now it looks good. – Jun 28 '13 at 11:51
Ah, "Now"! Fixed. – Cameron Buie Jun 28 '13 at 12:07

Convergence of $\sum_{n=0}^\infty z^{2^n}$

2 Answers2