For what values of the quotient $q>1$ the following property holds:
Every $x\in \mathbb R$ can be represented as $ x = \sum_{n=k}^\infty \pm q^{-n} \tag{*}\label{x} $ for some $k\in \mathbb Z$.
The case of $q=2$ is discussed here: Every real number can be represented as a sum of plus-minus of the terms of infinite geometric sequence $2^{-n}$.
No $q>2$ has the property as then $x=0$ can not be represented in the form \eqref{x}. Indeed, suppose that \eqref{x} represents $x=0$, then by triangle inequality \begin{align*} 0 = |x| &= \left| \sum_{n=k}^\infty \pm q^{-n} \right| \\ &\geq q^{-k} - \left| \sum_{n=k+1}^\infty \pm q^{-n} \right| \\ &\geq q^{-k} - \sum_{n=k+1}^\infty q^{-n} \\ &= q^{-k} - \frac{q^{-k-1}}{1-q^{-1}} \\ &= q^{-k} \left( 1 - \frac{1}{q-1} \right)\\ &= q^{-k} \cdot \frac{q-2}{q-1} > 0. \end{align*} Contradiction.
What can we tell about the case $q\in (1,2)$?
We first show that if $q\in (1,2]$ then every $x\in [-1,1]$ can be written in the desired way.
Fix $q\in (1,2]$ and $x_0 \in [-1,1]$ and consider the sequence $\{x_n\}$ given recursively by $$ x_n = \min_{a_n \in \{-1,1\}} |x_{n-1}-a q^{-n}|, \quad n \in \mathbb N. $$
Proof. We prove the lemma by mathematical induction:
which is the claim for $n$. $\tag*{$\Box$}$
We conclude by observing that $x_n \to 0$ as $n\to \infty$, and so $$ \sum_{n=1}^\infty a_n q^{-n} = x_0, $$ which is the desired representation of $x_0$.
Finally, given any $x\in \mathbb R$, let $k\in \mathbb Z$ be any number such that $|x|\leq q^{1-k}$. Then we put $x_0=q^{k-1}x$ and consider the sequence $\{a_n\}$ representing $x_0$. We have $$ x = q^{k-1} x_0 = q^{k-1} \sum_{n=1}^\infty a_n q^{-n} = \sum_{n=1}^\infty a_n q^{-n-k+1} = \sum_{n=k}^\infty a_{n-k+1} q^{n}, $$ which is the desired representation of $x$.