= Solution
Fix the <Riemann curvature tensor> convention
$$
R^\alpha{}_{\beta\mu\nu}=\partial_\mu\Gamma^\alpha{}_{\nu\beta}-\partial_\nu\Gamma^\alpha{}_{\mu\beta}+\Gamma^\alpha{}_{\mu\lambda}\Gamma^\lambda{}_{\nu\beta}-\Gamma^\alpha{}_{\nu\lambda}\Gamma^\lambda{}_{\mu\beta},\qquad R_{\beta\nu}=R^\alpha{}_{\beta\alpha\nu}.
$$
For the limiting <affine connection>, every product of two nonzero connection coefficients is zero, because its only nonzero coefficients have spatial upper index and both lower indices $t$. Its only possibly nonzero <curvature> components, apart from antisymmetry in the last two indices, are
$$
R^i{}_{tjt}=\partial_j\partial_i\phi.
$$
The <Ricci tensor> is therefore
$$
\boxed{\operatorname{Ric}(\nabla^{(\infty)})=(\Delta\phi)\,dt\otimes dt.}
$$
All other components vanish, so the limiting connection is a connection with vanishing <Ricci tensor> exactly when $\phi$ satisfies the <Laplace equation> on spatial <Euclidean space>. This is Ricci-flatness, not necessarily vanishing curvature: $\phi=x^2-y^2$ is a <harmonic function> but has nonzero <Hessian matrix> and hence a nonflat limiting connection. Reversing the curvature sign convention reverses the displayed Ricci sign but leaves the equivalence unchanged.
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