Suppose one has a decimal expansion of a real number $b:$ $$b=b_0.b_1b_2b_3...b_n...$$
where $b_n\in\mathbb{Z}$ and $0\leq b_n\leq9$ for $n\geq1$.
For which $b$ does $(b_n)$ converge?
Is the answer just real numbers that have period $0$ or $1$ decimal representations?
The idea behind this being since for $(b_n)$ to converge to $L$, we need, $\forall\epsilon>0, $ $|b_n-L|<\epsilon$ for $n$ greater than some $N\in\mathbb{N}$. Since $b_n$ consists of positive integer singletons, we need the sequence $(b_n)$ to be eventually constant; else, take $\epsilon:=\min\big\{\frac{b_i-b_j}{2}:0\leq i,j\leq9\big\}$. Then, no matter now large $N$ is, $|b_n-L|>\epsilon$ $\implies$ $(b_n)$ diverges.
Thus, so long as $b$ has a terminating decimal or eventually constant (one can view a terminating decimal as having infinitely many zeros at the end...) decimal expansion $(b_n)$, $(b_n)$ will converge.
Am I missing anything?
You're absolutely right that the eventually you need $b_n = b_{n+1} = \cdots$ since each $b_n$ can only be selected from $0,1,\cdots,9$. So either you're repeating $0$'s or repeating one of $1$ through $9$'s.
Lets assume for simplicity that $b_0 = 0$.
If you're repeating $0$'s, then $b = \frac{N}{10^n}$ for some $N <10^n$ and $n$ is the last nonzero digit.
If you're repeating $k$'s for $k \in \{ 1 \cdots 9\}$, then $b = \frac{N}{10^n} + \frac{k}{9 \times 10^{n}}$ where $n$ is the last nonrepeating digit.
For example, $0.58777777... = \frac{58}{100} + \frac{7}{900}$.
Note that this does not guarantee that $\frac{N}{10^n}$ or $\frac{k}{9 \times 10^{n}}$ can't be simplified further.
So, you could write $(b_n)$ converges if and only if there exists $n \in \mathbb{N}$, $N \in \mathbb{N}_0$ satisfying $N<10^n $, and $k \in \{0, \cdots 9\}$ such that $b - b_0 = \frac{N}{10^n} + \frac{k}{9 \times 10^{n}}$ .