What is the formal way of going from $(b_0, b_0+b_1p, b_0+b_1p+b_2p^2, \dots) \in \mathbb Z_p$ to the formal sum $b_0 + b_1 p + b_2 p^2 + \cdots$? I've figured out how to write an element of the $p$-adic integers $(a_1, a_2, a_3, \dots) \in \mathbb Z_p=\varprojlim \mathbb Z/p^i\mathbb Z$ as
$$(b_0, b_0+b_1p, b_0+b_1p+b_2p^2, \dots)$$ by considering base $p$ expansion for each least residue $a_i$.
However, I need to show that every element of $\mathbb Z_p$ can in fact be written as an infinite formal sum $$b_0+b_1p+b_2p^2+\cdots.$$
Looking at $(b_0, b_0+b_1p, b_0+b_1p+b_2p^2, \dots)$ it seems plausible.
But what is the formal way to state/write this?
 A: If you're given the task of showing that ever element of $\mathbb{Z}_p$ can be written as one of these formal sums, I think the bulk of the proof is in showing what you've already done in the first part of your question. From there, I don't think it takes any justification beyond something like:
"We can write $b\in\mathbb{Z}_p$ uniquely as a formal sum $b_0+b_1p+b_2p^2+\dots$ since truncating the sum at the $i$th term will give the $i$th component of the $p$-adic expansion of $p$."
edit for clarity: Basically you just want to show that the data contained in the $p$-adic expansion $(b_0,b_0+b_1p,b_0+b_1p+b_2p^2,\dots)$ can be recovered from the formal sum $b_0+b_1p+b_2p^2+\dots$ (and vice versa). If truncating the formal sum gives the individual components to the expansion, then these two things contain the same information.
A: For $b \in \mathbf Z_p$, show $|b - (b_0 + b_1p + \cdots + b_kp^k)|_p \leq 1/p^{k+1}$ for each $k$. Therefore, by the definition of $p$-adic limits, $b$ is the limit of $\sum_{j=0}^k b_jp^j$, which is what the equation $b = \sum_{j \geq 0} b_jp^j$ in $\mathbf Z_p$ means.
Example: $1/(1-p) - (1 + p + \cdots + p^k) = 1/(1-p) - (p^{k+1}-1)/(p-1) = p^{k+1}/(1-p)$, so $|1/(1-p) - \sum_{j=0}^k p^j|_p = 1/p^{k+1}$ for all $k \geq 0$. Therefore, letting $k \to \infty$, $1/(1-p) = \sum_{j \geq 0} p^j$.
