Let $p\_n$ denote the $n$-th consecutive prime number and $g\_n=p\_{n+1}-p\_n$ a prime gap. There are many results about the upper bound for $g\_n$. Some of them still has astatus of conjecture, such as Firoozbakth conjecture (in a prime gap version):
$g\_n<p\_n\left( \sqrt[n]{p\_n}-1\right) $ , $\forall n\in N$, and its consequence $g\_n<\sqrt{n}$ , $n\geq 3645$ . Currently the best known proved result on upper bound is $g\_n\leq p\_n^{0.525}$ ( Baker-Harman-Pintz Theorem ). Heuristic and a few calculations that I made, suggest for upper bound:


**Conjecture**:
$$\frac{g\_n}{\log{g\_n}}<2\log{n}, n\geq 5 $$


If we combine Conjecture with B.K.P. Theorem we get $g\_n<1.05\log^2{n}$ , with Firoozbakth Conjecture we get $g\_n<\log^2{n}$, which seem to be in contradiction with A.Granville proposition :$\limsup\_{n\longrightarrow\infty}\frac{g\_n}{\log^2{p\_n}}\geq 2e^{-\gamma}\approx 1.1229$. There is also a simple consequence of Conjecture providing the lower bound for a prime gaps:   

**Proposition**
$$g\_n>\left( \frac{p\_{n+1}}{p\_n}\right) ^{\frac{n}{2}} , n\geq 2$$   

Proof:
$$\left( \frac{p\_{n+1}}{p\_n}\right) ^{\frac{n}{2}}=\left[ \left( 1+\frac{g\_n}{p\_n}\right) ^{\frac{p\_n}{g\_n}}\right] ^{\frac{n.g\_n}{2p\_n}}<e^{\frac{n.g\_n}{2p\_n}}<e^{\log g\_n}=g\_n$$
( follow from Conjecture and $\frac{p\_n}{n}>\log n$) .   



**Question:** Is conjecture plausible? Can it be proved or disproved? If Conjecture is not true, can we prove the Proposition , now taken as Conjecture, with other means?



The lower bound for prime gaps
**1.** The Proposition in the post is almost equivalent to the Conjecture, namely it implies that
$$\frac{g\_n}{\log g\_n}\leq (2+o(1))\log n.$$
In particular, the Proposition (hence also the Conjecture) implies that
$$g\_n\ll\log n\,\log\log n.\tag{$\ast$}$$
This would contradict the common expectation that $g\_n\gg (\log n)^2$ holds for infinitely many $n$'s.


**2.** On the other hand, we don't know that $(\ast)$ is false. The best result in this direction is due to [Ford-Green-Konyagin-Maynard-Tao (2014)](https://arxiv.org/abs/1412.5029), and it states that
$$g\_n\gg\frac{\log n \,\log \log n\,\log\log\log\log n}{\log \log \log n}$$
holds for infinitely many $n$'s.


