Every real number can be represented as a sum of plus-minus of the terms of infinite geometric sequence $2^{-n}$.

Question

Every $C\in \mathbb R$ can be represented as $ C = \sum_{n=k}^\infty \pm 2^{-n} $ for some $k\in \mathbb Z$.

Trivially, every real number can be represented as $\sum_{n=1}^\infty \pm 2^{-a_n}$ for some strictly increasing sequence $\{a_n\}$. However, there might be missing terms in the geometric sequence, i.e. $a_{n+1}$ might be greater than $a_n$ for some $n$.

Is there a shorted / easier to follow proof than I provide among answers?

This question was inspired by the post: Conjecture about the representation of a constant $C=0.6516...$

Answer 1

Somewhat shorter proof:

Using the same argument as in the other proof, we can assume $C>0$. Let $\sum_{n=k}^\infty b_n 2^{-n}$, where $k\in \mathbb Z$ and $\{b_n\}\in \{0,1\}^\infty$, be the binary representation of $C$. Then $$ C = \sum_{n=k}^\infty b_n 2^{-n} = \sum_{n=k}^\infty 2b_n 2^{-(n+1)} = 2^{-k} + \sum_{n=k}^\infty (2b_n-1) 2^{-(n+1)}, $$ where $2b_n-1 = \pm 1$ for all $n=k,k+1,\ldots$. This is the desired representation.

Answer 2

Strategy: Notice that $$ 2^{-2} = 2^{-1} - 2^{-2}, $$ using this fact you can always "fill the gaps" in the sequence $\{a_n\}$. It allows me to prove a much stronger result in fact.

Proof.

If $C = 0$, then $C = 2^0 - \sum_{n=1}^\infty 2^{-n}$. It is enough to show that $C>0$ can be represented in the desired way, as in order to represent $-C$ it is enough to switch the signs.

In the rest of the proof we consider $C>0$. Since every positive real number can be represented in basis 2, there exist a (possibly finite) strictly increasing sequence $\{b_n\}$ such that $C = \sum_{n=1}^\infty 2^{-b_n}$. If the sequence was finite and $b_M$ was the last term, we could replace $2^{-a_M}$ by $$ \sum_{m=1}^\infty 2^{-(a_M+m)} = 2^{-a_M}. $$

We conclude that there is at least one way to represent $C$ as $\sum_{n=1}^\infty \pm 2^{-c_n}$, where $\{c_n\}$ is a strictly increasing sequence (with terms possibly apart by more than $10$). Let $\mathcal{S}$ be the set of all such sequences $\{c_n\}$. If $\mathcal{S}$ contains an arithmetic sequence of the form $c_1,c_1+1,\ldots$, we are done. Suppose that not. For a sequence $\{c_n\} \in \mathcal{S}$ define $$ f(\{c_n\}) = \min\big\{n\in \mathbb N:c_{n+1}-c_n>1\big\}. $$ and let $$ N = \max\big\{ f(\{c_n\}) : \{c_n\} \in \mathcal{S}\big\}. $$ Define the subset of $\mathcal{S}$ containing the sequence of "earliest occurance of a gap", $$ \mathcal{T} = \big\{ \{c_n\} \in \mathcal{F} : f(\{c_n\}) = N \big\}. $$ Finally, pick any sequence $\{c_n\}$ from $\mathcal{T}$ for which $c_{N+1}-c_N$ is minimal. By the definition of $N$ and $\mathcal{T}$, $c_{N+1}-c_N > 1$. By replacing $\pm 2^{-c_{N+1}}$ with $$ \pm 2^{-(c_{N+1}-1)} \mp 2^{-c_{N+1}}, $$ we obtain new sequence $\{c'_n\}$ that can be written as $c_1,\ldots,c_n,c_{N+1}-1,c_{N+1},\ldots$ and it belong to $\mathcal S$. If $c_{N+1}-c_{N}=2$, then the beginning of the new sequence contains more than $N$ consecutive integers ($f(\{c'_n)>N$), which is a contradiction with the maximality of $N$. Last, if $c_{N+1}-c_{N}>2$, then the new sequence $\{c'_n\}$ belongs to $\mathcal T$ and $c'_{N+1}-c'_{N} <c_{N+1}-c_{N}$, which is a contradiction with the way $\{c_n\}$ was chosen out of all the sequences in $\mathcal T$. That concludes the proof.