Hello,

this question is moreorless related to http://math.stackexchange.com/questions/57069/a-limit-involving-prime-numbers.

For $N$ a positive integer, let's define the set $A(N)$ as $\{m, \ \ \forall n\geq N, \ \ g_{n}:=(p_{n+1}-p_{n})<(\log p_{n})^{m}\}$. Let's assume that for all positive integer $H$, $A(H)$ is a non empty set. Then for any given positive integer $N$, $A(N)$ admits an infimum $m_{0}(N)$.

Let's furthermore assume Andrica's conjecture. This conjecture states that $\forall n>0, \ \ g_{n}<2\sqrt{p_{n}}+1$.

Due to the fact that $m_{0}(N)$ is the infimum of $A(N)$, there exists an integer $k\geq N$ such that $g_{k}=(\log p_{k})^{m_{0}(N)}$.
Now, suppose that $\forall n\geq N, \ \ (\log p_{n})^{m_{0}(N)}<2\sqrt{p_{n}}+1$.
Equivalently, $\forall x\geq p_{N}, \ \ (\log x)^{m_{0}(N)}\lt 2\sqrt{x}+1$.

So $\forall x\geq p_{N}, \ \ \log ((\log x)^{m_{0}(N)})\lt \log (2\sqrt{x}+1)$.`

Equivalently $\forall x\geq p_{N}, \ \ m_{0}(N)\log \log x\lt \log 2+\log (\sqrt{x}+1/2)$ and finally $\forall x\geq p_{N}, \ \ m_{0}(N)\lt \frac{\log 2+\log (\sqrt{x}+1/2)}{\log \log x}$.

As $m_{0}(N)$ doesn't depend on $x$ this means that $\forall x\geq p_{N},\ \ \frac{\log 2+\log (\sqrt{x}+1/2)}{\log \log x}\gt m_{0}(N)$. So one has $m_{0}(N)\lt y_{0}$, where $y_{0}=1.98772705...$ is the minimum of the function $f:x\gt e\mapsto \dfrac{\log 2+\log(\sqrt{x}+1/2)}{\log\log x}$, provided $p_{N}\lt x_{0}$ (where $x_{0}$ is such that $y_{0}=f(x_{0})$). As $67\lt x_{0}\lt 68$, one gets $N\leq 19$.

Furthermore, as $g_{n}\geq 2$, one must have $(\log p_{n})^{m_{0}(N)}\geq 2$. Since $m_{0}(N)>1$, one can take $n$ such that $\log p_{n}\geq 2$, hence $n\geq 5$. So that if for all integer $N$ $A(N)$ is a non empty set, and if Andrica's conjecture holds true, then (a stronger form of) Cramer's conjecture holds true, namely $c=0$, with $c$ defined in the link given at the beginning of this post.

Edit: I think I took Gerhard Paseman and quid's comments into account. So, is the proof correct now? Thank you in advance.