Interpreting big O for Sum of Primes < n The sum of primes < n has been answered (though not asked, some book/paper references of this would be nice too) by @Eric Naslund here:
What is the sum of the prime numbers up to a prime number $n$?
as
$$\sum_{p\leq x } \text{li}\left(x^{2}\right)+O\left(x^{2}e^{-c\sqrt{\log x}}\right),$$
I can see that $$\text{li}\left(x^{2}\right)$$ can be approximated using $$\operatorname{li}(x)\sim\frac{x}{\log(x)-1}$$ 
in which case as $$\ log(x^{2})=2log(x) $$
$$\operatorname{li}(x^{2})\sim\frac{x^{2}}{2log(x)}$$ 
so I can at least calculate numerically the first part correctly, but I am having difficulty with the big O part.
1) I want to calculate the big O numerically but,
2) I don't know what the c is or what to numerically put there.
3) Is the big O the maximum error (absolute) as the sum goes to infinity? or the growth rate of the error?
4) How would it calculate for me the error at say n=1000?
 A: By $O(f(x))$ we mean a function which, when divided by $f(x)$, gives a function with some non-zero finite limit as $x\to\infty$. (There are also problems where the intended meaning is as $x\to\infty$, e.g. $\ln (1+x)=O(x)$.) Technically $O(f(x))$ is defined as the set of such functions, so that $g=O(f)$ is an abuse of notation meaning $g\in O(f)$; you can interpret $g=h+O(f)$ similarly.
For numerical estimates you may be out of luck, because the term will approximate $Ax^2e^{-c\sqrt{\log x}}$ for some nonzero $A,\,c$. You might be able to prove some bounds on this from the logic in this answer. However, for an order-of-magnitude estimate it's usually safe to assume $A\approx 1$.
A: The prime number theorem $$\pi(x) = \sum_{2 \le n \le x} \left(\frac{1}{\log n}+\mathcal{O}( e^{-c\sqrt{\log n}})\right)$$ implies by partial summation
$$\sum_{p \le x} p =\sum_{2 \le n \le x} n \left(\frac{1}{\log n}+\mathcal{O}( e^{-c\sqrt{\log n}})\right)= \int_2^x \frac{t}{\log t}dt+\mathcal{O}(x^2 e^{-c\sqrt{\log x}})\\=\int_4^{x^2} \frac{\sqrt{u}}{\log \sqrt{u}}\frac{du}{2 \sqrt{u}}+\mathcal{O}(x^2 e^{-c\sqrt{\log x}})=\text{Li}(x^2)+\mathcal{O}(x^2 e^{-c\sqrt{\log x}})$$
And $\text{Li}(x)=\int_2^x \frac{dt}{\log t} \sim \frac{x}{\log x}$ follows from $\frac{x}{\log x} = \frac{2}{\log 2}+\int_2^x\left( \frac{1}{\log t}-\frac{1}{\log^2 t}\right)dt$
