This is a cross-post of this question on MSE. I would not usually do this, but have decided to in this case since it has had no responses having been posted as a bounty question. I did not delete the MSE question since so many users have starred it (and are apparently interested in seeing it answered) and not all of those users will have an MO account. If it receives a positive response here, I intend to link it to the MSE question for the benefit of those users. If however, this is objected to by MO users, or is not deemed 'research level', I shall respectfully remove it.

The question is: Is it likely that

$$\left|\operatorname{li}(n)-\sum_{k=1}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}-\log(2)-\dfrac{1}{2}\right|<\dfrac{2\sqrt{n}}{e\log(n)}?$$

This is of course, almost identical to saying

$$|R(n)-\pi(n)|<\dfrac{2\sqrt{n}}{e\log(n)}$$ where $R$ is the Riemann prime counting function, but this is a little too tight since this doesn't hold for $n=113$. Since

$$ \sum_{k=1}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}\approx\operatorname{li}(n)-\sum_{k=1}^{\infty}2\ \Re\left(\operatorname{Ei}\left(\rho_k\log\left(n\right)\right)\right)-\log(2) $$

and since $\pm\dfrac{2\sqrt{n}}{\Im(\rho_1)\log(n)}$ bounds $2\ \Re\left(\operatorname{Ei}\left(\rho_1\log\left(n\right)\right)\right)$

does it follow that $\pm\dfrac{2\sqrt{n}}{C\log(n)}$ will bound $\sum_{k=1}^{\infty}2\ \Re\left(\operatorname{Ei}\left(\rho_k\log\left(n\right)\right)\right)$ for some $C$ (assuming RH)?

*$C=e$ seems particularly tight.* Is this a realistic bound to propose?