I want to prove that $O\left(\hat{h}_n^{2(\bar{s}-1)} \exp \left(2 / \hat{h}_n^{\bar{s}}\right)\right)=O\left(h_0^{\bar{s}-1} \exp \left(2 / h_0^{\bar{s}}\right)\right)$,
where $\hat{h}_n =\left(\frac{\log n}{2}-\frac{\bar{\beta}-\hat{s}_n+1 / 2}{\hat{s}_n} \log \log n\right)^{-1 / \hat{s}_n}$ with $0<\underline s\le \hat{s}_n\le \bar s\le 2$ and $h_0=(\log n / 2)^{-1 / \underline s}$.
My attempt:
One can easily prove that $\hat{h}_n^2=O(\log(n/2)^{-\frac{1}{\underline s}})$. Therefore \begin{align*} \frac{\hat{h}_n^{2(\bar{s}-1)} \exp \left(2 / \hat{h}_n^{\bar{s}}\right)}{h_0^{\bar{s}-1} \exp \left(2 / h_0^{\bar{s}}\right)}&=\left(\frac{\hat{h}_n^2}{h_0}\right)^{\bar{s}-1}\exp \left(2 / \hat{h}_n^{\bar{s}}-2 / h_0^{\bar{s}}\right)\\ &\le\left(\frac{\hat{h}_n^2}{h_0}\right)^{\bar{s}-1}\exp \left(2 \log(n/2)^{\frac{\bar s}{\underline{s}}}-2 \log(n/2)^{\frac{\bar s}{\underline{s}}}\right)\\ &\le\left(\log( n / 2)\right)^{(1-\bar s) / \underline s}\overset{?}{<}\infty. \end{align*} Since I have $\hat{h}_n^2$ instead of $\hat{h}_n$ I'm left with $\log(n/2)$ which goes to infinity.