Let $n\in\mathbb{N}$, $k=\mathbb{F}_{p^{n}}$, $S=\operatorname{Spec}(k[t])$, $F$ the Frobenius automorphism on $S$ and $\mathcal{E}$ a sheaf of modules over $X$ such that $\mathcal{E}_{0}$ is of finite type.
Grothendieck's trace formula says that \begin{equation*}\sum_{s\in |S|^{F^{n}}}\operatorname{Trace}(F^{n},\mathcal{E}_{s})=\sum_{i=0}^{\infty}(-1)^{i+1}\operatorname{Trace}(F^{n},H^{i}_{c}(S,\mathcal{E}))\text{.} \end{equation*}
I wonder what happens if I replace $k[t]$ with $k[[t]]$, namely $\mathbb F_{p^n}[[t]]$?
What I have tried so far: A naive approach would be to consider $k[[t]]$ as the inverse limit of the rings $k[t]/t^{m}$, but this fails. I guess the reason for this is that each of the rings $k[t]/t^{m}$ has a spectrum consisting of a single point (the generic point and the unique maximal ideal coincide), whereas $\operatorname{Spec}(k[[t]])$ has two points (the generic point and the maximal ideal are different).
Question: Can the Grothendieck trace formula be modified in such a way that it works on $\operatorname{Spec}(k[[t]])$?