I just noticed that dividing $1 \div 998$ gives me the apparently non-periodic
$$0.001002004008016032064\ldots ,$$
which is $$10^{-3} + 2\times 10^{-6} + 4\times10^{-9} + 8\times 10^{-12} + \cdots = \sum_{i=0}^\infty 2^i 10^{-3(i+1)}.$$
Since every rational number expands into a finite or periodic decimal expansion, does that become periodic somehow? If so, how?
It does repeat, but with a very long period: Factoring the denominator into primes gives $$998 = 2 \cdot \color{#3f3fff}{499};$$ since $2$ is a factor of $10$ and $10$ is a primitive root modulo $\color{#3f3fff}{499}$, the period of repetition is $\color{#3f3fff}{499} - 1 = 498$ digits long. Indeed, consulting WolframAlpha gives:
$$\color{#bf0000}{ \begin{align} \smash{\textstyle\frac{1}{998}} = 0.&0\overline{0100200400801603206412825651302605210420841683366733466}\\ &\overline{93386773547094188376753507014028056112224448897795591182}\\ &\overline{36472945891783567134268537074148296593186372745490981963}\\ &\overline{92785571142284569138276553106212424849699398797595190380}\\ &\overline{76152304609218436873747494989979959919839679358717434869}\\ &\overline{73947895791583166332665330661322645290581162324649298597}\\ &\overline{19438877755511022044088176352705410821643286573146292585}\\ &\overline{17034068136272545090180360721442885771543086172344689378}\\ &\overline{757515030060120240480961923847695390781563126252505}. \end{align} }$$
A more manageable example of this phenomenon is the repetition of the decimal digits of, e.g., $\frac{1}{14}$: Since $14 = 2 \cdot 7$ and $10$ is a primitive root modulo $7$, the period of the decimal expansion of $\frac{1}{14}$ is $7 - 1 = 6$: $\frac{1}{14} = 0.0\overline{714285}$.