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Not a very well known inequality is $$p_{n+1}-p_n\leq n$$ where $p_i$ is the $i^{th}$ prime number.

I know this can be proven using the following inequality: (B. Rosser, L. Schoenfeld) $$\forall x\geq 67 \text{ we have }\frac{x}{\log x-\frac{1}{2}}<\pi(x)$$

and proposition $6.8$ of this article of P. Dussart.

However, I am interested if anyone knows a simpler proof. In particular, one that does not use approximations of $\psi(x)$ and $\vartheta(x)$ presented in the article I linked above.

Thank you!

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    Possibly helpful: https://math.stackexchange.com/q/981267/42969 – Martin R Apr 09 '21 at 16:26
  • I already read that thread and the response suggests the same solution I know. Thanks for posting it though! –  Apr 09 '21 at 17:07
  • it's implied by Grimm's conjecture – Roddy MacPhee Apr 09 '21 at 20:20
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    Not convinced there is a non-PNT proof. $\pi(x)=Li(x)(1+O(1/\log^2 x))$ gives $p_n=Li^{-1}(n)(1+O(1/\log^2 n)), p_{n+1}-p_n = O(n/\log n)$ so it remains to check finitely many $n$. – reuns Apr 10 '21 at 00:52
  • @reuns are you sure that your argument is enough for a general proof? Indeed, you proved that this is true for $n$ greater than an arbitrary constant $c$ but you cannot find $c$, so I do not see how you can prove this for any $n$. –  Apr 11 '21 at 08:27
  • The PNT gives an effective $O$ constant thus some $c$. – reuns Apr 11 '21 at 09:42
  • I am not sure you understood what I meant. Anyways, thanks for the suggestion. –  Apr 11 '21 at 19:31

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