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Bertrand's Postulate gives us that:

$$p_n < p_{n+1} < 2p_n$$

So that:

$$p_{n+1} - p_n < p_n$$

In this answer, this paper is cited which says in Prop 6.8 that:

For $x \ge 396738$ there is always a prime in the interval $[x, x + x/(25\ln^2 x)]$

Is this the best upper bound known for consecutive primes?

If I understand it correctly, does it mean that:

$$p_{n+1} - p_n < \frac{p_n}{25\ln^2 p_n}$$

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The best result is a result of Baker, Harman and Pintz (Baker, R. C.; Harman, G.; Pintz, J. (2001) "The difference between consecutive primes, II". Proceedings of the London Mathematical Society 83 (3): 532–562) which states that $$p_{n+1}-p_{n}<p_{n}^{\theta}$$ with $\theta=0.525$ and a sufficiently large $n.$

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  • $\begingroup$ Is the lower bound of $n$ known? $\endgroup$ – Larry Freeman Mar 22 '15 at 21:43
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    $\begingroup$ @LarryFreeman The main theorem states that the inequality holds "for all $x>x_{0}$", so I think it is not computed the lower bound of $n$. $\endgroup$ – Marco Cantarini Mar 23 '15 at 16:20
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    $\begingroup$ @LarryFreeman: I've been told that the proof is effective, in principle, but as far as I know it hasn't been computed. $\endgroup$ – Charles Mar 23 '15 at 17:03

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