p-adic expansion of a rational number

As @ThomasAndrews says, you don’t need to rearrange anything. First, knock off the nonperiodic part at the beginning. That will be a rational number. Now take the periodic part, say of period $N$. Then the part you didn’t knock off has the form $A + Ap^N + Ap^{2N} +Ap^{3N}+\cdots$, a geometric series with common ratio $p^N$. Since this ratio is $p$-adically smaller than $1$, the series is convergent, and the periodic part has rational value $A/(1-p^N)$, done.

A $\,p-$adic series converges iff its general term sequence converges to zero (this is every undergraduate freshman's dream!) , so in this case the series's convergence is automatic since

$$a_np^n\xrightarrow [n\to\infty]{}0$$

The periodicity of the coefficients $\,a_n\,$ now give us that the limit must be rational (and re-arranging the series terms is possible since the convergence here is absolute: what is true for the usual absolute value is true for the $\,p-$adic value as well...and sometimes things get nicer in the latter).

In this case, you are really grouping the sums into finite sub-sums. So if you know that $\sum_{k=n}^\infty b_k$ exists, then given any sequence of increasing natural numbers $n=n_0<n_1<\dots$ you can define $c_i = \sum_{k=n_i}^{n_{i+1}-1} b_k$ and get $\sum_{i=0}^\infty c_i = \sum_{k=n}^\infty b_k$. This is all you need to prove that the eventually-repeating $p$-adic number is rational.

You do essentially the same thing - indeed, it is exactly the same proof - to prove that any real number with repeating decimal expansion is rational.

It is not true that if $\sum c_i$ us convergent then $\sum b_k$ exists. But in $p$-adic numbers, we know $\sum b_k$ exists if and only if $b_k\to 0$. Proving this varies, depending on how you define $p$-adic numbers.