Given $\varepsilon>0$ there are only a finite number of primes $p_n$ such that $p_{n+1}-p_n>p_n^{\,\varepsilon}$.

Do anyone have an idea if this can be proved or not?

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The method used is to test all primes $< 100,000,000$ and calculate the number of hits and the largest prime in the set of hits.

$\quad\varepsilon\quad\quad\quad\quad p_{max}\quad\quad hits$

0.50           113        6  
0.40        31,397       41  
0.30    47,326,693    2,003  
0.25    99,988,649   54,726 

For $\;\varepsilon=0.25\;$ is so close to the limit ($100,000,000$) that $p_{max}$ and $hits$ probably are much greater than in the diagram.


Quotes and most of the information from https://en.wikipedia.org/wiki/Prime_gap#Upper_bounds :

"Bertrand's postulate, proved in 1852, states that there is always a prime number between $k$ and $2k$, so in particular $p_{n+1} < 2p_n$, which means $g_n < p_n$."

"Hoheisel (1930) was the first to show that there exists a constant $\theta < 1$ such that

$\pi (x+x^{\theta })-\pi (x)\sim {\frac {x^{\theta }}{\log(x)}}{\text{ as }}x\to \infty ,$

hence showing that $g_n<p_n^\theta,$ for sufficiently large $n$."

Clearly, with the first quote, for any $\theta > 1$, the count is 0. As to the second quote, we see your statement holds as we go down to $\theta = 0.525$. As for $0 < \theta < 0.525$, we have no proof that this is the last prime counted for that $\theta$ because stronger result is computationally hard (as seen with the web page). But on the way, we know that gaps between some primes get arbitrarily large and the prime count increases. At $0_+ = \theta$, the count is $\infty$. The graph, like your graph, will make you think of the graph of $1/x$ but with the floor function, or $1/c\lfloor c/x\rfloor$.

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