An inequality on prime gaps 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 $p_{k}-p_{k-1}$?
 A: 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 $2e^{-\gamma}\approx1.1229\ldots.$
A: 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
A: 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.
