A limit involving prime numbers Me and a friend of mine worked on building a problem for AMM. It all started pretty well, but in the end we realized that the initial part of the solution was wrong. In few words, we thought we have proven that 

$$\lim_{n \to \infty} \left(\log_{p_{n+1}} ((n+1)!)-\log_{p_n}(n!)\right)=1$$

where $p_n$ is the $n$-th prime number.
Denoting $x_n=\log_{p_n}(n!)$, there are a few ways that I think it is possible to prove that $x_{n+1}-x_n \to 1$:
First, maybe it is possible by direct computation (we tried and didn't get anything). The second method is by using Stolz-Cesàro in a different way: if we prove that $(x_{n+1}-x_n)$ is convergent then it should have the same limit as $x_n/n$ which converges to $1$.
So my question is:

Is it true that $(x_{n+1}-x_n) \to 1$, or at least $(x_{n+1}-x_n)$ is convergent?

Thank you.
 A: I am going to assume that Harald Cramer's conjecture holds true, i.e. $g_n = p_{n+1}-p_n = \mathcal{O}(\log^2 p_n)$. 
Let's further assume that the limit $\displaystyle\lim_{n\to \infty} \frac{g_n}{\log^2 p_n}$ exists and let $\displaystyle c = \lim_{n\to \infty} \frac{g_n}{\log^2 p_n}$ denote its value. Then $\displaystyle\lim_{n\to\infty} x_{n+1}-x_{n} = 1-c.$
To show this simplify the difference by using the factorial's recurrence equation:
$$ 
   x_{n+1}-x_n = \frac{\log p_n \cdot  \log (n+1) - \log n! \cdot \log(1+\frac{g_n}{p_n})}{ \left(\log p_n + \log(1+\frac{g_n}{p_n}) \right) \log p_n }
$$
Now use $\displaystyle\lim_{n\to \infty} \frac{\log n!}{p_n} = 1$. And write
$$
  x_{n+1}-x_n = \frac{\log p_n \cdot  \log (n+1) - \frac{\log n!}{p_n} \cdot p_n \log(1+\frac{g_n}{p_n})}{ \left(1 + \frac{1}{\log p_n} \log(1+\frac{g_n}{p_n}) \right) \log^2 p_n }
$$
And, finally:
$$
  x_{n+1}-x_n = \frac{ \frac{\log (n+1)}{\log p_n} - \frac{\log n!}{p_n} \cdot \frac{g_n}{\log^2 p_n} \cdot \frac{p_n}{g_n} \log(1+\frac{g_n}{p_n})}{ \left(1 + \frac{1}{\log p_n} \log(1+\frac{g_n}{p_n}) \right) }
$$
Now, since $\displaystyle\lim_{n\to\infty} \frac{g_n}{p_n} = 0$, $$\displaystyle\lim_{n\to\infty} \frac{p_n}{g_n} \log \left(1+\frac{g_n}{p_n}\right) = 1$$ and $$\displaystyle\lim_{n\to\infty} \left(1 + \frac{1}{\log p_n} \log \left(1+\frac{g_n}{p_n}\right) \right) = 0.$$
A: Let $m_{0}$ be the infimum of the set $\{m\in \mathbb{N} : \forall n \text{ sufficiently large, }  g_{n}<(\log p_{n})^{m}\}$. Then one can prove than $m_{0}$ is not an integer. Thus it is sufficient to prove that $m_{0}<2$ to get $c=0$ and the desired limit.
