maybe this is a stupid question, but I could not solve it even for the ordinary integers $\mathbb{Z}$. Furthermore, I don't have to much knowledge on algebraic number theory and ramifications.
Let $K$ be a number field and $v$ a discrete valuation on it. Now let $\mathcal{O}_{K,v}$ be the localization at $v$, with maximal ideal $\mathfrak{p}_v = \mathfrak{p}\mathcal{O}_{K,v}$. Is it true that $\widehat{\mathcal{O}_{K,v}} \cong \varprojlim_n \; \mathcal{O}_K/\mathfrak{p}^n$ ?
On
In general, if $\mathfrak m$ is a maximal ideal in the commutative ring $A$, then the natural map from the $\mathfrak m$-adic completion of $A$ to the $\mathfrak m$-adic completion of the localization $A_{\mathfrak m}$ is an isomorphism, the reason being that at each finite stage, the morphism $$A/\mathfrak m^n \to A_{\mathfrak m}/\mathfrak m^n A_{\mathfrak m}$$ is an isomorphism, because any element of $A\setminus \mathfrak m$ has invertible image in $A/\mathfrak m$, and hence in $A/\mathfrak m^n$.
This last assertion is where we use that $\mathfrak m$ is maximal, and the statement is not true if $\mathfrak m$ is not maximal.
Yes, both are profinite completions and the residue fields are isomorphic.
That is
$$\mathcal{O}_{K,v}/\mathfrak{p}^n\cong \mathcal{O}/\mathfrak{p}^n$$
where each $\mathfrak{p}$ is within the ambient ring, so that the inverse limits are isomorphic, since the profinite completion of $\mathcal{O}_K$ WRT $\mathfrak{p}$ is given over the inverse system of progressive quotients by powers of $\mathfrak{p}$ as is true for the localizations.
In particular, the inverse systems are
and there is a componentwise isomorphism which is compatible with the transition maps (each arrow is reduction modulo the next lowest prime power), hence is an isomorphism of inverse systems.