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Let $p_k$ be the $k$-th prime.

Then $\lim_{n\to\infty}\frac{p_{n+1}}{p_n}=1$ -- this is well known.

I was looking for more specific information: What is the exponent of convergence of $\frac{p_{n+1}}{p_n}$ to $1$, and found no answer.

I mean, what is $\lim_{n\to\infty}n^a(\frac{p_{n+1}}{p_n}-1)$ in terms of $a$ (of course $a>0$).

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    $\begingroup$ Finding the $a$ such that your limit converges to $0$ is exactly the prime gap problem. The RH implies $a < 1/2$ works $\endgroup$ – reuns Apr 5 at 19:16
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This is an open question so the right answer is we don\t know yet. But we can look at some of the well know conjecture on the ratio (or gap) of two consecutive primes.

The strongest conjecture we have in this regards is due to Reza Farhadian.

$$ p_n^{\big(\frac{p_{n+1}}{p_n}\big)^n} \le n^{p_n} $$

A corollary of this conjecture is the weaker Firoozbakht conjecture $p_{n+1}^{1/n+1} < p_n^{1/n}$ which has been verified for $n \le 10^{17}$. Both these conjectures are however believed to be false in light of the Cramer-Granville heuristics.

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