Consider the partial sum $S_n = \sum_{i=0}^n (a_i \bmod 2) \cdot a_i \cdot 2^i$ and observe that $\lim_{n \to \infty} S_n$ converges in $\mathbb{Z}_2$ to a dyadic integer for all $a_i \in \mathbb{Z}$ where infinitely-many are odd. (Most sources define $p$-adic series where the coefficients are in $\{0,\dots,p-1\}$, and I believe one reason is because it ensures that the assignments are unique.) My question is the following:
$\textbf{Question:}$ Does there exist a sequence of positive, distinct integers $\{a_i\}_{i=0}^{\infty}$ and a positive integer $m$ such that $$\sum_{i=0}^\infty (a_i \bmod 2) \cdot a_i \cdot 2^i = -m$$ in $\mathbb{Z}_2$?
If this can be resolved, consider the same problem but where (1) $a_i \in \mathbb{Z}$ and (2) $a_i \in \mathbb{Z}_+$ and for the more general equation $$\sum_{i=0}^\infty a_i \cdot 2^i = -m.$$
Thank you very much in advance.
Sure. Indeed, you can find such a series that converges to any desired $\alpha\in\mathbb{Z}_2$ at all (for either your original question or for variant (2)). Suppose that so far you have chosen $a_0,\dots,a_{n-1}$ so that $S_{n-1}=\alpha$ mod $2^n$. To choose the next $a_n$ so that $S_n=\alpha$ mod $2^{n+1}$, all that matters is the parity of $a_n$. So, you can choose whatever value of $a_n$ you want of the correct parity; in particular, you could choose $a_n$ to be a positive integer that is distinct from $a_0,\dots,a_{n-1}$. In the end you obtain a sequence $(a_i)$ such that $S_n=\alpha$ mod $2^n$ for each $n$, so that $S_n$ converges to $\alpha$.
(And of course nothing is special about $2$ here; you can make analogous statements for $\mathbb{Z}_p$ for any $p$.)