# Partial sums of $\sum_0^\infty z^n$

Let $$z$$ be a complex number with $$|z|<1$$. For every subset $$A\subset\mathbb N$$, the series $$\sum_{m\in A}z^m$$ is convergent. Denote $$S(A)\in\mathbb{C}$$ its sum and $$\Sigma_z$$ the set of all numbers $$S(A)$$. Remark that the cardinal of $$\Sigma_z$$ is (likely) that of $${\cal P}({\mathbb N})$$, the continuum.

Is it possible that $$\Sigma_z$$ be a neighbourhood of the origin ?

• The question is: Does there exist $|z|<1$ such that $\Sigma_z=\{\sum\limits_{n\in A} z^n: A\subseteq \mathbb N_0\}$ is a neighbourhood of $0$. – Jochen Wengenroth May 27 at 8:00
• @JochenWengenroth Why is $\alpha\notin\mathbb Q$ necessary when $r<1$? – Emil Jeřábek May 27 at 9:14
• To record what I just understood from @JochenWengenroth's comment: $r \geq 1/2$ is necessary because $\Sigma = z\Sigma \cup (1 + z\Sigma)$ and therefore $\mu(\Sigma) \leq 2r\mu(\Sigma)$ where $\mu$ is Lebesgue measure. ($\Sigma$ is evidently compact and therefore of finite measure.) – François G. Dorais May 27 at 9:28
• I agree with @EmilJeřábek. Wouldn't $z = i/\sqrt{2}$ work knowing that every integer has a negabinary representation? mathworld.wolfram.com/Negabinary.html – François G. Dorais May 27 at 9:43
• @FrançoisG.Dorais $\mu(z\Sigma) = r^2 \mu(\Sigma)$, right? So that gives $r \geq 1/\sqrt 2$. – Bart Michels May 27 at 12:18

The number $$z = i/\sqrt2$$ seems to work!

Given $$x \in [-2/3,4/3]$$ we can find a "negabinary" expansion $$x = \sum_{k=0}^\infty (-1)^k\frac{b_k}{2^k},$$ where each $$b_k \in \{0,1\}$$. Similarly, given $$y \in [-2/3\sqrt2,4/3\sqrt2]$$ we can find $$y = \frac{1}{\sqrt2}\sum_{j=0}^\infty (-1)^j\frac{c_j}{2^j},$$ where each $$c_j \in \{0,1\}$$. Therefore, $$x + iy = \sum_{n \in A} z^n$$ where $$A = \{2k : b_k = 1\} \cup \{2j+1 : c_j = 1 \}.$$

As explained in comment contributions by Bart Michels, Jochen Wegenroth and myself, $$|z| \geq 1/\sqrt2$$ is necessary. By definition, $$\Sigma_z = z\Sigma_z\cup(1+z\Sigma_z).$$ If $$\mu$$ denotes Lebesgue measure, then $$\mu(z\Sigma_z) = |z|^2\mu(\Sigma_z)$$ thus $$\mu(\Sigma_z) \leq 2|z|^2\mu(\Sigma_z)$$. Since $$\Sigma_z$$ is compact, it has finite measure and thus if $$\mu(\Sigma_z)>0$$ then we must have $$|z|^2 \geq 1/2$$.

It remains open whether $$|z|\geq1/\sqrt2$$ and $$z \notin \mathbb{R}$$ is sufficient for $$\Sigma_z$$ to contain $$0$$ in its interior.

• I'm a bit tired so the interval bounds might be a bit off... Please check my arithmetic! – François G. Dorais May 27 at 10:10
• Denis: thank you for correcting my accidental abuse of the letter $i$. For what its worth, I think the case $z= i/\sqrt2$ is anecdotal and doesn't fully answer your question. – François G. Dorais May 27 at 11:02
• François, it answers my question. Of course, it raises the more general question of which z has this property ?. By the way, how do you prove that $\Sigma_z$ is measurable ? – Denis Serre May 27 at 13:38
• Denis, $\Sigma_z$ is compact since it is a continuous image of $2^{\mathbb N}$. Therefore it is measurable and has finite measure. – François G. Dorais May 27 at 16:05
• See Davis, Chandler; Knuth, Donald E. Number representations and dragon curves-I. J. Recreational Math. 3 (1970), no. 2, 66–81. and part II in J. Recreational Math. 3 (1970), no. 3, 133–149. – Gerald Edgar May 27 at 16:26