By prime number theorem, it is possible to shown that $$ \liminf_{n\to\infty}{p_{n+1}-p_n\over\log p_n}\le1, $$ and I learned from an exercise in Cojocaru & Murty that there exists $0<\delta<1$ such that $$ \liminf_{n\to\infty}{p_{n+1}-p_n\over\log p_n}\le1-\delta. $$ The authors of the book hint that this is achieved using a proof by contradiction (i.e. assuming $p_{n+1}-p_n\ge(1-\delta)\log p_n$ for all large $n$):
Let $q_1,q_2,q_t$ denote all primes in $[x,2x]$. It is trivial that $$ S=\sum_{1\le k\le t-1}(q_{k+1}-q_k)\le x. $$ Then the authors say that breaking the sum according whether $q_{k+1}-q_k\in I=[(1-\delta)\log x,(1+\delta)\log x]$ or not. Consequently we have $$ S=\sum_{\substack{q_{k+1}-q_k\in I}}(q_{k+1}-q_k)+\sum_{\substack{q_{k+1}-q_k\notin I}}(q_{k+1}-q_k). $$ Using Brun's sieve I am able to show that the first sum is $\ll\delta x$. This suggests that there exists a constant $A>0$ such that for large $x$ $$ S\ge(1+\delta)\log x\sum_{\substack{q_{k+1}-q_k\notin I}}1-A\delta x. $$ I don't know how to handle the remaining sum, so I wonder whether anyone in this community can help me complete the proof of this result.