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Is it possible that

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

enter image description here

Since

$$ \sum_{k=1}^{\lfloor\log(n)\rfloor}\dfrac{\pi(n^{1/k})}{k}\approx\operatorname{li}(n)-\sum_{k=1}^{\infty}2\ \Re\left(\operatorname{Ei}\left(\rho_k\log\left(n\right)\right)\right)-\log(2) $$ (where $\rho_k$ is $k$th zeta zero) shown as a partial sum below

enter image description here

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

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.

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 $n=113$.

Note

The log plot is particularly striking:

enter image description here

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I will start by making it clear that this is not my answer.

Since this question generated a fair amount of interest and had not been answered, I thought it best to cross-post on MO & then link back for the benefit of those who starred it.

Here is the link to GH from MO's answer.

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    $\begingroup$ Maybe should mark it CW then...+1 for your question! $\endgroup$ – draks ... Sep 8 '14 at 8:02

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