Fix $d>0$, and let $n$ be a positive integer going to infinity, and for $h>0$, define $E=E(n,h)$ by
$$ E(n,h) := h\sqrt{\log \frac{1}{h}} + \frac{1}{h}\sqrt{\frac{d\log n}{n}}. $$
A strategy $h=h_n$ for selecting $h$ as a function of $n$ is called admissible, if the following conditions are respected:
- (i) $h_n \to 0$ in the limit $n \to \infty$, and
- (ii) $E(n,h_n) \to 0$ in the limit $n \to \infty$.
Question. What is a selection strategy $h=h_n$ which ensures that $E(n,h_n) \to 0$ as fast as possible ? In particular, is there a selection strategy for which $E$ decays asymptotically as rapidly as $1/n^{\alpha}$ (ignoring log factors), for some $\alpha \ge 1/2$ ?
Example
Just to be sure we're not searching in a vacuum, consider the following strategy $$ h_n := \left(\dfrac{\log n}{n}\right)^s, \tag{1} $$ for $s > 0$, independent of $n$. $$ \begin{split} E(n, (1/n)\log n) &= s\left(\dfrac{\log n}{n}\right)^s (\sqrt{\log n} - \log\log n) + \left(\dfrac{n}{\log n}\right)^s\sqrt{\dfrac{d\log n}{n}}\\ & \asymp s\frac{(\log n)^{1/2+s}}{n^s} + \sqrt{d}\left(\frac{\log n}{n}\right)^{1/2-s} \end{split} $$
Thus, the strategy (1) is admissible if and only if $s \in (0,1/2)$.
Moreover, it is clear that the value of $s \in (0,1)$ which makes $E$ decrease to zero fastest is $s=1/2$. For this value of $s$, $E$ decreases to $0$ at the same rate as $n^{-1/4}$ (ignoring log terms), which is short of the ultimate goal (namely $1/n^{1/2}$).