Given a graph $G$, we say $S$ is a dominating set if $S\cup \{N(x):x\in S\}=V(G)$. Let $d(n,k)$ be the smallest integer $s$ so that every $n$-vertex graph $G$ with minimum degree $k$ has some dominating set $S$ with $|S|\le s$.
By considering random $k$-regular graphs, one can show that $d(n,k)\ge (1-\epsilon_k)\ln(k)\frac{n}{k+1}-o(n)$, where $\epsilon_k\to 0$ as $k\to \infty$. As sketched from page 4 of https://arxiv.org/pdf/0810.2053.pdf, it seems that taking $\epsilon_k = (3+\delta_k)\frac{\ln\ln(k)}{\ln(k)}$ is permissible (where $\delta_k\to 0$ as $k\to \infty$). Thus, $$d(n,k)\ge \Big(\ln k-O(\ln\ln k)\Big)\frac{n}{k+1}.$$
Meanwhile, the best upper bound I am aware of is $$d(n,k)\le \Big(\ln k+1-o_k(1)\Big)\frac{n}{k+1}$$ (the term $\ln(k+1)+1$ is obtained in Theorem 1.2.2 of The Probabilistic Method, but you can improve the this by some decaying additive amount by skipping a simplifying approximation).
I am wondering if there are any sharper bounds known. I would assume that the lower bound is the truth. Is it open whether $$d(n,k)\le \Big(\ln k-10^{10}\Big)\frac{n}{k+1}+O(1)$$ for large $k$?
