Asymptotic bounds on $\pi^{-1}(x)$ (inverse prime counting function)

Assuming the Riemann hypothesis $$|p_n -\text{li}^{-1}(n) |\le \pi^{-1} \sqrt{n} (\log n)^{5/2}, \qquad n \ge 11$$ This follows from known bounds for $\pi(x)$ due to Schoenfeld.

You may see the details in the paper http://front.math.ucdavis.edu/1203.5413

Here is how we invert the prime counting function $\pi(x)$ to estimate the n-th prime $p_n$. Let $g(u),u \ge 2$ be positive, continuous increasing function and let $f(x)$ be defined by

$$ x = \int_{2}^{f(x)} \frac{du}{g(u)} $$

Then

$$ f(x) = \int_{2}^{x} g(f(u))du + f(2) $$

Take $g(u) = \ln u$ so our first estimate of $f(x) is $that $f(x) \sim x\ln x$. Now use this estimate in the above integral to obtain the second estimate for $f(x)$. You can repeat this iterative process $k$ times to obtain the asymptotic expansion of $f(x)$ with an error $O(x/(\ln x)^{-k})$.

Now from the Prime number theorem, we have

$$ \pi(x) \sim \int_{2}^{x} \frac{du}{\ln(u)} $$

Taking $x = p_n$, the n-th prime, the above method says that the asymptotic expansion of $p_n$ is $f(n)$. Thus we obtain

$$ p_n \sim n\ln n + n\ln\ln n - n + \ldots $$

Since "the state of the art" slightly changed since 2013, I am adding this new result on lower and upper bounds on $p_n$. (This is complementary to the answers given earlier.)

Christian Axler proved in https://arxiv.org/abs/1706.03651 that:

(1) For every $n\ge46254381$, $${\small p_n < n \Big(\log n + \log\log n - 1 + {\log\log n - 2\over\log n} - {(\log\log n)^2 - 6\log\log n + 10.667 \over 2\log^2 n} \Big). } $$

(2) For every $n\ge2$, $${\small p_n > n \Big(\log n + \log\log n - 1 + {\log\log n - 2\over\log n} - {(\log\log n)^2 - 6\log\log n + 11.508 \over 2\log^2 n} \Big). } $$