Prove or disprove that for every positive integer $k$, there exist three consecutive primes $p_{i-1}, p_i, p_{i+1}$ such that $p_i-p_{i-1}\gt k$ and $p_{i+1}-p_{i}\gt k$.
It's well known that $n!+2,n!+3,\cdots,n!+n$ are composite numbers for any integer $n\gt1,$ therefore, there exist gaps between primes which are arbitrarily large. Another way to see that arbitrarily large prime gaps must exist is the fact that the density of primes approaches zero, according to the prime number theorem.
I think we may prove the statement by contradiction. If it were not true, then there exist an integer $k$ such that for every prime $p$ there exist a prime $q$ such that $\left|{p-q}\right|\le k.$ Therefore,
$$\sum_{p,q\in \mathbb P,\left|{p-q}\right|\le k}\left(\frac{1}p+\frac{1}q\right)\gt \sum_{p \in \mathbb P}\frac{1}p=\infty.\tag 1$$
Therefore, if we can prove that $(1)$ is not true, namely, $$\sum_{p,q\in \mathbb P,\left|{p-q}\right|\le k}\left(\frac{1}p+\frac{1}q\right)\lt \infty,$$ then we will complete our proof.
This is very similar to Brun's theorem, which states that the sum of the reciprocals of the twin primes (pairs of prime numbers which differ by 2) converges to a finite value known as Brun's constant,
$$\sum_{p,p+2 \in \mathbb P}\left(\frac{1}p+\frac{1}{p+2}\right)\lt \infty.$$
There maybe exist a direct and simple proof, for example, we may use the prime number theorem.
Thanks in advance!
Consider the congruences $x\equiv1\pmod2,x\equiv2\pmod3,\dots,x\equiv k\pmod{p_k}$ together with $x\equiv-1\pmod{p_{k+1}},x\equiv-2\pmod{p_{k+2}},\dots,x\equiv-k\pmod{p_{2k}}$ where $p_r$ is the $r$th prime. By the Chinese Remainder Theorem, these are equivalent to a single congruence $x\equiv M\pmod P$ for some $M$, where $P$ is the product of the first $2k$ primes. By Dirichlet's Theorem on Primes in Arithmetic Progression, there are infinitely many primes $p$ satisfying this congruence. But for any such prime $p$, the numbers $p+1,p+2,\dots,p+k$ are all composite, as are the numbers $p-1,p-2,\dots,p-k$.