Given $x \in [0, 1]$, we denote by $e(x)$ the complex number $e^{2 \pi i x}$.
Can we characterise the set of rationals $x$ for which the sum
$$A_N(x)\, :=\, \sum_{n = 0}^N e(2^n x)$$
remains bounded as $N \to \infty$?
Remarks:
We allow the bound to depend on $x$, but not $N$.
It can be shown that the sum is unbounded for almost every (but not every!) irrational $x$.
As Fedor Petrov expects in the comments, this is true for $x=a/b$ if and only if the order of $2$ modulo $b$ is greater than the order of $2$ modulo the product of prime factors of $b$.
To prove this, the first key observation is given by Peter Mueller in the comments: Since the sequence is eventually periodic, it is bounded if and only if the sum over a period is $0$. We can also just drop the first few terms by multiplying $x$ by a power of $2$ until $b$ is odd, in which case the sequence is literally periodic. So it is equivalent to say $$\sum_{n=0}^{m-1} e( 2^n a/b) =0 $$ where $m$ is the multiplicative order of $2$ modulo $b$.
The next key observation is from GH from MO in the comments, which is that $\sum_{n=0}^{m-1} e( 2^n a/b) $ is a Galois conjugate of $\sum_{n=0}^{m-1} e( 2^n a'/b) $ whenever $a'$ and $a$ are both relatively prime to $b$, so one of them is zero if and only if all of these are zero, i.e. if and only if $$\sum_{n=0}^{m-1} e( 2^n a'/b) =0 \textrm{ for all } a' \in (\mathbb Z/b\mathbb Z)^\times$$
Now finally I will make a contribution of my own: By Fourier analysis, we have $$\sum_{n=0}^{m-1} e( 2^n a'/b) =0 \textrm{ for all } a' \in (\mathbb Z/b\mathbb Z)^\times$$ if and only if $$ \sum_{c \in (\mathbb Z/b\mathbb Z)^\times} e( c/b) \chi(c) =0 \textrm{ for all } \chi\colon (\mathbb Z/b\mathbb Z)^\times \to \mathbb C^\times \textrm{ such that } \chi(2)=1.$$
Indeed the first sums depend on the equivalence class of $a'$ in $(\mathbb Z/b\mathbb Z)^\times/\langle 2\rangle$, and the second sums give the finite Fourier transform over this group.
Now the theory of Gauss sums tells us that $\sum_{c \in (\mathbb Z/b\mathbb Z)^\times} e( c/b) \chi(c) =0 $ if and only if $\chi$ factors through $(\mathbb Z/(b/p)\mathbb{Z})^\times$ for some $p$ with $p^2$ dividing $b$.
If the order of $2$ modulo $b$ is greater than the order of $2$ modulo the product of prime factors of $p$, raising $2$ to the order modulo the product of prime factors and then raising to a suitable product of prime factors, we can find a power of $2$ that is congruent to $1$ mod $b$ but not modulo $b/p$ for some such prime, showing that every $\chi$ admits such a factorization. Conversely, if these orders are equal, then every character of the kernel of $(\mathbb Z/ b \mathbb Z)^\times \to (\mathbb Z/r \mathbb Z)^\times$, with $r$ the product of prime factors, extends to a character of $(\mathbb Z/ b \mathbb Z)^\times$ trivial on $2$. The kernel is cyclic and extending a faithful character produces a nonzero Gauss sum.
Not a full answer, but here is a potentially useful observation - let $x$ be a real number, rational or otherwise. By writing $x$ in binary, truncating to a certain number of binary digits to the right of the decimal point, and then counting the error terms, we can show that
$$A_N (x) \leq o(1) + \inf_{k \geq 2 \log_2 N} \sum_{n = 1}^{N} e(T_{k}(2^n x)).$$
where $T_m$ denotes the truncation to $m$ binary digits operator, and the $o(1)$ denotes an error term independent of $x$ that goes to $0$ in absolute value as $N \to \infty$.
This formula allows to get very precise estimates on the sum. Indeed, since $x \to 2^n x$ is just the shift map in binary, we can use this formula to show that in fact, for the rational number $\frac{10}{99} = 0.10101010\dots$, the sum $A_N (\frac{10}{99})$ is in fact bounded as $N \to \infty$ (see here for proof!), and we may even obtain the exact explicit bound if desired.
The formula may also be useful for checking boundedness on irrationals.
