Asymptotic bounds on $\pi^{-1}(x)$ (inverse prime counting function) What are the current best asymptotic bounds on $\pi^{-1}(x)$, where $\pi(x)$ denotes the prime counting function (number of primes at most $x$)?
In other words, I am curious about the state of the art for estimating the $n^{th}$ prime. From the prime number theorem, it seems clear that $\pi^{-1}(x)=\Theta(x \log x)$. Can someone point me in the direction of literature that answers the question? Please excuse my inability to find such literature myself...
 A: 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 https://arxiv.org/abs/1203.5413
A: 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
$$
p_n \sim n\ln n + n\ln\ln n - n + \ldots
$$
A: 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 $n\ge2$,
$${\small
p_n > n \Big(\log n + \log\log n - 1 + {\log\log n - 2\over\log n} - {(\log\log n)^2 - 6\log\log n + 11.508 \over 2\log^2 n} \Big).
}
$$
A: [I think this is morally the same as Nilotpal Sinha's answer, but it only uses asymptotic estimates and no integrals, so I thought I'd post it here in case someone found it useful or interesting.]
Start with the version of the Prime Number Theorem which says that $\pi(x) \sim x / \ln(x)$. Let $p_n$ denote the $n$-th prime, so in particular $\pi(p_n) = n$, and substitute in $p_n$ in the asymptotic estimate for $\pi(x)$ to get $n = \pi(p_n) \sim p_n / \ln(p_n)$, or equivalently
$p_n \sim n \ln p_n$
(If you're being nitpicky this works because the sequence $p_n$ is monotonic and unbounded.) Now take logs of both sides (again, if you're being nitpicky, this works because both $n$ and $p_n / \ln(p_n)$ are unbounded) and multiply by $n$ to get:
$n \ln p_n \sim n \ln(n) + n\ln\ln(p_n)$.
At this point, if you want just the leading term, you can show easily that $\lim_{n \to \infty} n \ln\ln(p_n) / n\ln(n) = 0$, so that in fact $p_n \sim n \ln p_n \sim n \ln n$, or you can probably keep substituting in to get finer and finer error terms.
