# Bound for $|p_n - \operatorname{li}^{-1}(n)|$

It is well-known that $$|\pi(x)-\operatorname{li}(x)| \leq \epsilon(x)$$, where $$\pi(x) = \sum \limits_{p \leq x} 1$$ is the prime counting function, where $$\operatorname{li}(x) = \int \limits_{2}^{x}\frac{1}{\ln(t)} dt$$ is the logarithmic integral, and where the function $$\epsilon$$ satisfies $$\lim \limits_{x\to\infty} \frac{\epsilon(x)}{\operatorname{li}(x)} = 0$$. My question is as follows:

Is anything known about an upper bound for $$|p_n - \operatorname{li}^{-1}(n)|$$? Here $$p_n$$ is the $$n$$-th prime.

• I was reading online about asymptotically $n$-th prime number by JUAN ARIAS DE REYNA, and i found this in the paper : if $\pi(x) = li(x)+O(\epsilon(x))$ then $p_x = li^{-1}(x) + O(\epsilon(x \ln(x)) \ln(x))$. I think this is what i am looking for, but i don't understand the $O$ notation, could any one give me precise answer. – Ahmad Mar 22 '17 at 21:04
• In analytic number theory, $O(f(x))$ denotes any function whose absolute value is less than some constant times $|f(x)|$. So if you have a good approximation $\pi(x)\approx\mathrm{li}(x)$, then you have a good approximation $p_n\approx\mathrm{li}^{-1}(n)$. It is straightforward to work out the relation of the two $O$-constants in the result you quote. – GH from MO Mar 22 '17 at 21:52