prime numbers' upper bound involving Lambert W [proof]

Question

let $p_{n}$ be the nth-prime number and $W_{t}$ the Lambert-W function / $W(z)e^{W(z)}=z $ /, show that:

$ \exists \ \ l \in \mathbb N:\forall n>l$

$$ p_{n}<n-nW_{-1}\bigg(\frac{-e^{2}}{n}\bigg)$$

Answer 1

C. AXLER in [1] showed that $ \exists \ \ l \in \mathbb N:\forall n>l$ :

$ p_{n}<n\Big(\ln(n)+\ln_{2}(n)-1+\frac{\ln_{2}(n)-2}{\ln(n)}-\frac{\ln_{2}(n)^{2}-6\ln_{2}(n)+10.667}{2\ln(n)^{2}}\Big)=\Theta_{n}$ $ p_{n}>n\Big(\ln(n)+\ln_{2}(n)-1+\frac{\ln_{2}(n)-2}{\ln(n)}-\frac{\ln_{2}(n)^{2}-6\ln_{2}(n)+10.667}{2\ln(n)^{2}}\Big)=\Phi_{n} $

So

$p_{n}=n\Big(\ln(p_{n}-n)-(\ln(p_{n}-n)-\frac{p_{n}}{n})\Big)<n\Big(\ln(p_{n}-n)-(\ln(\Phi_{n} -n)-\frac{\Theta_{n}}{n})\Big)$

$j_n=\ln(\Phi_{n} -n)-\frac{\Theta_{n}}{n}$

Since we have $j_n>1 \ $for $n\geq32$ (by computation)

$p_{n}<n\Big(\ln(p_{n}-n)-j_n\Big)<n\Big(\ln(p_{n}-n)-1\Big)$.

$p_n+n<n\ln(p_{n}-n)$

$e^{\ \frac{p_n}{n}+1}<p_n-n$

$e^{2}e^{\ \frac{p_n}{n}-1}<n(\frac{p_n}{n}-1)$

$-\frac{e^{2}}{n}>-(\frac{p_n}{n}-1) e^{-(\ \frac{p_n}{n}-1)}$

$$ p_{n}<n-nW_{-1}\bigg(\frac{-e^{2}}{n}\bigg) $$

$ \ $

$ \ $

[1] NEW ESTIMATES FOR THE n-TH PRIME NUMBER, CHRISTIAN AXLER, Jun 2017 https://arxiv.org/pdf/1706.03651.pdf