As you have observed since $a_k$'s are digits from $0$ to $9$, it will converge only when there exists an $N$ such that $a_n = d \in \{0,1,2,3,4,5,6,7,8,9\}$ for all $n>N$.
(To see this, choose say $\epsilon = 0.1$. Since $a_n$ converges, this means there exists $N(\epsilon)$, such that $\vert a_{n+1} - a_n \vert < \epsilon$ for all $n > N$. But $a_n, a_{n+1} \in \{0,1,2,\ldots,9\}$. Hence, $a_{n+1} = a_n$.)
Hence, $a = a_0.a_1a_2\ldots a_N dddddd \ldots$ $10^Na = a_0 a_1 \ldots a_N . dddd\ldots$ $10^{N+1}a = a_0 a_1 \ldots a_N d. dddd\ldots$ Hence, $10^{N+1}a - 10^Na = (a_0 a_1 \ldots a_N d - a_0 a_1 \ldots a_N) = M$ $a = \dfrac{M}{9 \times 10^N}$ i.e. in the simplest form ($\gcd$ of numerator and denominator is $1$) the numbers are of the form $\dfrac{M}{2^{n_1}3^{n_2}5^{n_3}}$ where $n_1,n_3 \in \mathbb{N}$, $n_2 \in \{0,1,2\}$ and $M \in \mathbb{Z}$ such that $\gcd(M,2^{n_1}3^{n_2}5^{n_3}) = 1$.