Let $X_Z$ be the formal completion of $X$ along $Z$. Let's assume we are working in $char=p$. How does $K_i(Z)$ compare to $K_i(X_Z)$? You can assume everything is smooth and $Z$ is a prime divisor in $X$ defined by a single equation if necessary, you can also invert any primes or work with rational groups if required. By algebraic $K$-theory of formal completion I mean the $K$-theory of formal vector bundles. $X_Z^{(n)}$ is the $n$-th nilpotent thickening of $X$ along $Z$. According to Thomason we have the following result:

According to Theorem 9.7, let's say in char $p$, the pullback along the morphism $Z\rightarrow X_Z^{(n)}$ induces isomorphisms on $K$-groups after tensoring with $\mathbb{Z}[1/p]$.

There is a natural morphism $K_i(X_Z)\rightarrow \varprojlim K_i(X_Z^{(n)})$. The inverse limit is over $n$. When is this morphism injective? (maybe after inverting some primes like $p$.)

Q) For simplicity let's assume that all groups $K_i(X_Z^{(n)})$ are trivial for all $n$ and $i>0$. Does that imply the triviality of $K_i(X_Z)$? What can be said about $K_i(X_Z)$ in this case?

Edit: It seems to me that algebraic $K$-theory of formal vector bundles is not well-studied, $K$-theory of formal completion is usually defined as the inverse limit of $K$-theory of formal thickenings rather than $K$-theory of formal vector bundles. Are there any resources for the $K$-theory of the formal vector bundles?

A more general question could be how is the algebraic $K$-theory of inverse limit/limit of exact categories is related to each individual exact category? A somewhat relevant result is due to Carlsson that claims algebraic $K$-theory commutes with infinite products.

Here is a proof which I think it works for the rational version of Q) if everything is affine (Not sure whether it has a chance of getting generalized for non-affine case):

Edit2: I am assuming that in the affine case, higher $K$-theory of formal vector bundles can be given by the using the plus construction on $GL(\hat{A})$. But I am not sure if this is correct. This requires that formal vector bundles to be cofinal among trivial formal vector bundles.

Because we are working rationally we just need to work with the homology of general linear groups. Because of the assumption in Q) if $X=\text{Spec}(A)$ and $I$ be the ideal corresponding to $Z$, we have $H_i(GL(A/I^n),\mathbb{Q})=0$ for all $n$ and $i>0$. If $\hat{A}:= \text{lim}A/I^n$, we want to show $H_i(GL(\hat{A}),\mathbb{Q})=0$ for $i>0$. First thing to notice is that every element of $GL(A/I^n)$ can be lifted to an element of $M(A/I^{n+1})$ which automatically falls in $GL(A/I^{n+1})$, we call this the liftability property. Second fact is that $GL(\hat{A})=\text{lim}GL(A/I^n)$. Let $C_*^i$ be the chain complex corresponding to $GL(A/I^i)$. The chain complex of $GL(\hat{A})$ is given by $\text{lim}C_*^i$. Because of the liftability property at each degree the transition maps induce surjection on the chain complexes. So each degree of the chain complex $\text{lim}C_*^i$ is an inverse limit of abelian groups of the form $\ldots \rightarrow A_2\rightarrow A_1 \rightarrow A_0$ with surjective transition maps. Let $\partial$ be the operator defined here. Then because of surjectivity it implies that $\partial$ is surjective. So we get an exact sequence of the following form: $$0\rightarrow \text{lim}A_i \rightarrow \prod_iA_i \rightarrow \prod_iA_i \rightarrow 0$$

This implies that $\text{lim}C_*^i$ fits into a short exact sequence of the following form:

$$0\rightarrow \text{lim}C_*^i \rightarrow \prod_iC_*^i \rightarrow \prod_iC_*^i \rightarrow 0$$ Writing the homology long exact sequence for this and using the fact homology of $C_*^i$ are rationally trivial at positive degrees implies that homology of $\text{lim}C_*^i$ or equivalently $GL(\hat{A})$ are rationally trivial at positive degrees which finishes the proof.

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