What are the bases $\beta$ such that a number with non-periodic expansion can be approximated with infinitely many numbers with periodic expansion? Warning: This problem requires a bit of setting.
Fix a finite set $A \subset \Bbb{Z}$ and consider an infinite non (ultimately) periodic sequence $\mathbf{a}=(a_i)_{i \geq 1}$ of elements of $A$ such that there are a real number $w > 1$ and sequences $(U_n),(V_n)$ of words on $A$ such that (here $|W|$ is the length of a word $W$):


*

*For every index $n$ the word $U_nV_n^w$ is a prefix of $\mathbf{a}$, where $V^w = V^{\lfloor w\rfloor}V'$ and with $V'$ a prefix of $V$ of length $\lceil(w - \lfloor w\rfloor)|V| \rceil$

*The sequence $\left(\frac{|U_n|}{|V_n|}\right)$ is bounded from above

*The sequence $(|V_n|)$ is increasing
and for each index $n$ let $\mathbf{b}^{(n)}$ be the (periodic) sequence given by
$$
 \begin{cases}
    b_i^{(n)} = a_i & \text{for }\, 1 \leq i \leq r_n \\
    b_{r_n + i + hs_n}^{(n)} = a_{r_n + i} & \text{for }\, 1 \leq i \leq s_n
                                           \,\text{ and }\, h \geq 0
 \end{cases}
$$
where $r_n$ and $s_n$ are the lengths of $U_n$ and $V_n$, respectively.
Finally, fix an algebraic number $\beta \in \Bbb{C}$ with $|\beta| > 1$ (here $|\cdot|$ is the usual complex absolute value), define
$$
\alpha = \sum_{i = 1}^{\infty} \frac{a_i}{\beta^i}
\qquad
\alpha^{(n)} = \sum_{i = 1}^{\infty} \frac{b^{(n)}_i}{\beta^i}
$$
It can be shown that there is a constant $C > 0$ depending only on $A$ and $\beta$ such that
$$
|\alpha - \alpha^{(n)}| < C \beta^{-r_n -ws_n -1} \label{eq:1}\tag{1}
$$

Question: What are the most general $\beta$ such that \eqref{eq:1} implies that $\alpha^{(n)}$ admits infinitely many different values?
  For example, this is the case if $\beta$ is a Pisot number.

See also these two related questions:


*

*Can a number have both a periodic an a non-periodic representation in a non-integer base?

*Monograph about periodic representations of numbers in non-integer bases
Note: I encountered this problem while working on my master thesis, trying to extend the main result from this paper by B. Adamczewski and Y. Bugeaud.
 A: As a partial answer, we show that, for all but countably many positive real $\alpha$ and any real $\beta$ in the range $1<\beta\leqslant\frac32,$ there is $n\in\Bbb Z$ and continuum-many sequences $\pmb c=(c_0,c_1,...)\in \{0,1,2\}^\Bbb N$ such that $$\alpha=\sum_{i=0}^\infty c_i\beta^{n-i}.$$As a corollary, all but countably many of these sequences are aperiodic.
First choose $n\in\Bbb Z$ and $c_0\in\{0,1\}$ so that $c_0\beta^n<\alpha<(c_0+1)\beta^n$. Suppose that $c_0,...,c_{k-1}$ are given, and write $$\alpha_k:=\frac1{\beta^{n-k}}\left(\alpha-\sum_{i=0}^{k-1}c_i\beta^{n-i}\right)\quad\text{and}\quad\beta':=\frac{2-\beta}{\beta-1}.$$Now the ($k+1$)-tail of the expansion, namely $$\sum_{i=k+1}^\infty c_i\beta^{n-i}=\alpha-\sum_{i=0}^kc_i\beta^{n-i}=\alpha_k\beta^{n-k}-c_k\beta^{n-k},$$must be at most $$\sum_{i=k+1}^\infty\beta^{n-i}=(\beta'+1)\beta^{n-k},$$from which it follows that $c_k\geqslant\alpha_k-\beta'-1.$ Also, the ($k+1$)-term partial sum$$\sum_{i=0}^{k-1}c_i\beta^{n-i}+c_k\beta^{n-k}$$cannot exceed $\alpha$. From these two bounds we get$$\alpha_k-\beta'-1\leqslant c_k\leqslant\alpha_k.$$This interval for $c_k$ is guaranteed to contain two integers when $\beta'\geqslant1,$ namely when $1<\beta\leqslant\frac32.$ In the case when the greater one is $0$, we must choose it. As long as $\alpha$ is not one of the countably many numbers with a terminating expansion, namely of the form $\sum_{k=n}^mc_k\beta^k$, there will be be infinitely many places in the expansion where we can choose between two options, $0$ or $1$, for the digit, allowing continuum-many sequences $\pmb c.$
Remark: this includes the particular cases $\beta=\frac32$ and $\beta=\sqrt2$ given in answers to another question by the O.P.
Edit: The original version of this answer had a mistake in the proof. The proof has now been corrected to demonstrate a weaker result. It may be that the original result is true, but it needs a more complicated proof to deal with the (now excluded) case when $\alpha$ has a terminating expansion.
