# Is the $n$-th prime smaller than $n(\log n + \log\log n-1+\frac{\log\log n}{\log n})$?

Let $p_n$ be the $n$-th prime. Wikipedia gives the following known bounds on $p_n/n$ when $n\geq6$: $$\log n+\log\log n-1 \leq \frac{p_n}{n} \leq \log n+\log\log n.$$

If I take the first few terms in the asymptotic expansion for $p_n/n$, like so: $$\frac{p_n}{n} = \log n+\log\log n-1+\frac{\log\log n}{\log n} - \frac{2}{\log n} + O\left(\frac{(\log\log n)^2}{\log^2 n}\right),$$ it follows that $$\frac{p_n}{n} < \log n+\log\log n-1+\frac{\log\log n}{\log n} + \frac{c}{\log n},$$ for $c>-2$ and large enough $n$. For at least those $n$ that I've checked however ($p_n\leq 10^{11}$), I find that this inequality holds with $c=0$ and $p_n>347$, and also with $c=-1$ and $p_n > 5393$.

Is this actually correct?

Is there a sharper inequality?

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As a general warning, $10^{10}$ is a relatively small number in regards to conjectures like these. –  Hurkyl Jul 2 '13 at 5:19
Fair enough, I just ran out of memory. –  Kirill Jul 2 '13 at 5:25
A recent paper : de Reyna and Jeremy 'The n-th prime asymptotically'. –  Raymond Manzoni Jul 2 '13 at 23:37
@RaymondManzoni Thank you. I followed their reference to arxiv.org/abs/1002.0442, a paper by Pierre Dusart, which has the desired inequality on page 2. Can you submit your comment as an answer, mentioning the Dusart paper, so that I can accept it? –  Kirill Jul 3 '13 at 0:21
I am Glad the link was useful @Kirill and provided an answer (I didn't know then that Dusart's fine work was exactly the answer you wanted since it was 2am here...). –  Raymond Manzoni Jul 3 '13 at 10:05

Knowing more zeta zeros allowed indeed Pierre Dusart in his $1998$ thesis to update Rosser and Schoenfeld's earlier results and get new bounds on $\pi(x)$ (other results are here in french).
He extended his results to bounds for $p_n$ in his $1999$ paper 'The Kth prime is greater than k(ln(k)+ln(ln(k))-1) for k>=2' and in $2010$ in the paper you found 'Estimates of Some Functions Over Primes without R.H.' and which contains indeed the precise : $$\frac{p_n}n\le\ln(n)+\ln(\ln(n))-1+\frac{\ln(\ln(n))-2}{\ln(n)}\quad\text{for}\ n\ge 688383,$$ $$\frac{p_n}n\ge\ln(n)+\ln(\ln(n))-1+\frac{\ln(\ln(n))-21/10}{\ln(n)}\quad\text{for}\ n\ge 3.$$