Continuing from here, since
$$ \sum_{k=1}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}=\operatorname{li}(n)-\sum_{k=1}^{\infty}2\ \Re\left(\operatorname{Ei}\left(\rho_k\log\left(n\right)\right)\right)-\log(2)+\int_{n}^{\infty}\dfrac{\operatorname{d}t}{t(t^2-1)\log(t)} $$
and since
$$\left|2\ \Re\left(\operatorname{Ei}\left(\rho_1\log\left(n\right)\right)\right)\right|\leq\dfrac{2\sqrt{n}}{\Im(\rho_1)\log(n)},$$
surely, it follows that
$$\left|\sum_{k=1}^{\infty}2\ \Re\left(\operatorname{Ei}\left(\rho_k\log\left(n\right)\right)\right)\right|\leq\dfrac{2\sqrt{n}}{C\log(n)}$$
for some $C$ (assuming RH). If this is the case, then
$$\operatorname{li}(n)-\sum_{k=1}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}-\log(2)+\int_{n}^{\infty}\dfrac{\operatorname{d}t}{t(t^2-1)\log(t)}=O\left(\dfrac{\sqrt{n}}{\log(n)}\right)$$
Would it not then follow that $$\pi(n)=\operatorname{li}(n)+O\left(\dfrac{\sqrt{n}}{\log(n)}\right)?$$
Are these not the "best possible" prime bounds under the RH, rather than
$$\pi(n)=\operatorname{li}(n)+O\left(\sqrt{n}\log(n)\right)?$$