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It is not difficult to calculate upper bounds on $s(n)$ from bounds on the prime counting functions function $\pi(n)$. Just use integration by parts, $$s(n) = \int_0^n x\,d\pi(x) = n\pi(n) - \int_0^n\pi(x)\,dx.$$ I'm not sure what the currently best known bounds for $\pi(x)$ are but, checking Wikipedia, gives $$\frac{x}{\log x}\left(1+\frac{1}{\log x}\right) < \pi(x) < \frac{x}{\log x}\left(1+\frac{1}{\log x}+\frac{2.51}{(\log x)^2}\right)$$ with the left hand inequality holding for $x\ge599$ and the right hand holding for $x\ge355991$. So,

$$s(n)\le \frac{n^2}{\log n}\left(1+\frac{1}{\log n}+\frac{2.51}{(\log n)^2}\right)-\int^n\left(1+\frac{1}{\log x}\right)\frac{x\,dx}{\log x}+c$$ (where $c$ is a constant which you can compute if you feel so inclined). Applying integration by parts,

$$s(n)\le\frac{n^2}{2\log n}\left(1+\frac{1}{\log n}+\frac{5.02}{(\log n)^2}\right)-\frac12\int^n\left(1+\frac{2}{\log x}\right)\frac{x\,dx}{(\log x)^2}+c$$

Bounding $\log x\le\log n$ in the integral gives a bound

$$s(n)\le\frac{n^2}{2\log n}\left(1+\frac{1}{2\log n}+\frac{4.02}{(\log n)^2}\right)+c$$

You can also take $c=0$ if you only require the bound to hold for $n\ge N$ (some $N$), since the term I neglected in the integral by applying $\log x\le \log n$ grows withuot bound, and will eventually dominate any constant term. Obviously, if you know any better bounds for $\pi(n)$ then you will get improved bounds for $s(n)$. For example, the same Wikipedia article linked to above states that $\left\vert\pi(x)-{\rm Li}(x)\right\vert\le\frac{\sqrt{x}\log x}{8\pi}$ for $x\ge2657$ under the assumption that the Riemann hypothesis holds.

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It is not difficult to calculate upper bounds on $s(n)$ from bounds on the prime counting functions $\pi(n)$. Just use integration by parts, $$s(n) = \int_0^n x\,d\pi(x) = n\pi(n) - \int_0^n\pi(x)\,dx.$$ I'm not sure what the currently best known bounds for $\pi(x)$ are but, checking Wikipedia, gives $$\frac{x}{\log x}\left(1+\frac{1}{\log x}\right) < \pi(x) < \frac{x}{\log x}\left(1+\frac{1}{\log x}+\frac{2.51}{(\log x)^2}\right)$$ with the left hand inequality holding for $x\ge599$ and the right hand holding for $x\ge355991$. So,

$$s(n)\le \frac{n^2}{\log n}\left(1+\frac{1}{\log n}+\frac{2.51}{(\log n)^2}\right)-\int^n\left(1+\frac{1}{\log x}\right)\frac{x\,dx}{\log x}+c$$ (where $c$ is a constant which you can compute if you feel so inclined). Applying integration by parts,

$$s(n)\le\frac{n^2}{2\log n}\left(1+\frac{1}{\log n}+\frac{5.02}{(\log n)^2}\right)-\frac12\int^n\left(1+\frac{2}{\log x}\right)\frac{x\,dx}{(\log x)^2}+c$$

Bounding $\log x\le\log n$ in the integral gives a bound

$$s(n)\le\frac{n^2}{2\log n}\left(1+\frac{1}{2\log n}+\frac{4.02}{(\log n)^2}\right)+c$$

Obviously, if you know any better bounds for $\pi(n)$ then you will get improved bounds for $s(n)$. For example, the same Wikipedia article linked to above states that $\left\vert\pi(x)-{\rm Li}(x)\right\vert\le\frac{\sqrt{x}\log x}{8\pi}$ for $x\ge2657$ under the assumption that the Riemann hypothesis holds.