So I checked Method of finding a p-adic expansion to a rational number
and also P-adic expansion of rational number
Many questions here is of the form: Given a rational number, find its p-adic expansion. The solution to this is long division or some sort of geometric series argument.
What if I want to go backwards. i.e., given the p-adic expansion, what is the rational number assosciated to it? (Of course, not every p-adic expansion is rational so from reading texts, it seems the p-adic expansion of rational numbers eventually repeat so lets assume the ones we work with do)
To make this a concrete question, lets say I have like
$5^{-1} + 1 + 2*5 + 5^2 + (4*5^3 + 2*5^4 + 3*5^5 + 2*5^7 + 5^8) + (4*5^9 + 2*5^{10} + 3*5^{11} + 2*5^{13} + 5^{14}) + (4*5^{15} + 2*5^{16} + 3*5^{17} + 2*5^{19} + 5^{20})+...$
where I put the repeated parts in parenthesis. So if the first term is $5^{-1}$, lets say it is .1121(423021 repeat). How would I find the rational number associated to it?
I would write: $$ x = {\ldots}\overline{120324}121.1_5 $$ and then add/subtract the appropriate finite base-5 fraction to make it a pure repeating number ending at the ones: $$x-1.1_5 = {\ldots}\overline{120324}120_5 = {\ldots}\overline{324120}_5 $$ Then the I can convert that to a fraction essentially like one does for an ordinary decimal fraction, except that the analogue of $0.999{\ldots}_{10}=1$ is now ${\ldots}444_5=-1$: $$ \begin{align} x - \tfrac65 &= 324120_5 \cdot {\ldots}\overline{000001}_5 = \frac{324120_5}{444444_5} \cdot {\ldots}\overline{444444}_5 = -\frac{324120_5}{444444_5} \\ \implies\quad x &= \frac{6}{5} - \frac{324120_5}{444444_5} = \frac65 - \frac{11160}{15624} = \frac{37944}{78120} =\frac{17}{35} \end{align} $$
Alternatively you could start with $$ \frac{x - 121.1_5}{5^3} = {\ldots}\overline{120324}_5$$ and convert the right-hand side of that to a fraction in the same way.