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Sppose I have the following property: $ \forall \alpha \in (0,1) \hspace{0.5cm}\exists K_{\alpha} \in \mathbb{N} $ such that $ \forall k \geq K_{\alpha}$ one has $$\hspace{0.5cm}p_{n_k+1}^{\alpha} < p_{n_k}^{\alpha} + 2$$ where $p_{n_k}$ is a prime number. What can we conclude from here for $p_{n_k+1} - p_{n_k}$ ?

I tried for instance $\alpha = \frac{1}{2}$ then it comes out $\sqrt{p_{n_k+1}} < \sqrt{p_{n_k}} + 2$ hence $p_{n_k+1} - p_{n_k} < 4\sqrt{p_{n_k}} + 4$ I tried similarly for $\alpha = \frac{N-1}{N}$ with Newton binomial, but is a little too complicated ...

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  • $\begingroup$ I guess that $n_k\uparrow +\infty$. In this case, any $c$ satisfies the wanted condition as $(n_k+c)^\alpha-n_k^\alpha$ goes to zero as $k$ goes to infinity. $\endgroup$
    – Davide Giraudo
    Commented May 22, 2017 at 16:20

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If $a> 0$ and $b < 1$ then as $n \to \infty$ : $$(n+n^b)^a - n^a = \int_n^{n+n^b} a x^{a-1}dx \sim \int_n^{n+n^b} a n^{a-1}dx = a n^{a+b-1}$$ So $(n+n^b)^a - n^a < 2 \Leftrightarrow n^b < \frac{n^{1-a}}{a}$ for $n$ large enough.

In the same way, since $p_{k+1}-p_k = \mathcal{O}(p_k^{1-\epsilon})$ we obtain $$p_{k+1}^a-p_k^a < 2 \quad \Leftrightarrow \quad p_{k+1}-p_k < \frac{p_k^{1-a}}{a} \quad (\text{for } k \text{ large enough})$$

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  • $\begingroup$ I think I am able now (thanks to some ideas aroud here ... :) ) to prove that for each $\alpha \in (0,1)$ there is a sequence $(p_{n_k})_{k \in \mathbb{N}}$ depending on $\alpha$ such that $p_{n_k + 1}^{\alpha} - p_{n_k}^{\alpha} < 2$ ... that is (as you sir proved) $p_{n_k+1} - p_{n_k} < \frac{p_{n_k}^{1-\alpha}}{\alpha}$ Is this something new? $\endgroup$
    – C Marius
    Commented May 23, 2017 at 10:05
  • $\begingroup$ If $\alpha$ would be $\alpha = 1$ this is twin prime conjecture ... can I extend the this results some how, to prove the existence of a sequence of primes for $\alpha = 1$ given the fact that it exists for $\forall \alpha \in [0,1)$ ... ? $\endgroup$
    – C Marius
    Commented May 23, 2017 at 10:08

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