Is there a good inequality for prime gaps. Like $p_{k}-p_{k-1}\leq f(k)$ ? In other words is there a known upper bound for $p_{k}-p_{k-1}$?
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2I'm not sure but can't you get a good approximation using $\pi(x)$? – Guy Mar 30 '14 at 13:23
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$p_{n+1}-p_n<n$ according to this. – rtybase Sep 14 '19 at 11:21
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Bertrand's postulate gives that $p_k-p_{k-1}\le p_{k-1}.$ A result of Baker, Harman, & Pintz can be used to improve this to $p_k-p_{k-1}\ll p_{k-1}^{0.525}.$
It is conjectured that $p_k-p_{k-1}\ll \log^2 p_{k-1},$ perhaps with a constant as small as $2e^{-\gamma}\approx1.1229\ldots.$
Charles
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You can find some results in the following link: https://primegaps.files.wordpress.com/2016/09/gap-between-consecutive-primes-by-using-a-new-approach.pdf
Free.san
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2Though this may address the question, it is a good idea to include some essential points as part of your answer – Shailesh Sep 23 '16 at 03:36
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The result in this paper appears to be of the same strength Cramér's conjecture. Given how recent it is, and how short it is without any discussion of any major ideas, I am very skeptical of its correctness. – Erick Wong Sep 23 '16 at 16:51
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Let $g_n=p_{n+1}-p_n$. Consider the series $\sum_{p\leq x} g_n/p_n$. By Mertens theorem, one obtains
$\dfrac{g_{n+1}}{p_{n+1}} = 2\log \big(\dfrac{\log p_{n+1}}{\log p_n}\big)$,
ignoring an error term that vanishes as $n \to \infty$. By the PNT, there exists $\varepsilon>0$ depending on $x_0 \in \mathbb{N}$ such that for $p_n>x_0$ we always find $p_{n+1} \leq (1+\varepsilon)p_n$. Hence, we can deduce that
$g_{n+1} \leq C_{\varepsilon}\dfrac{p_{n+1}}{\log p_n}$,
for some constant $C_{\varepsilon}$ depending on $\varepsilon$, with $0<C_{\varepsilon}<1$. In line with this thread, can anyone check if this argument holds.
Buzzman
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