# Confusion on the proof of Erdös's prime gap inequality

I am currently reading Erdös's paper "The difference of consecutive primes" published in 1940, in which he shows that there exists $$\delta>0$$ such that $$A=\liminf_{n\to\infty}{p_{n+1}-p_n\over\log p_n}\le1-\delta$$ His method is essentially a proof by contradiction:

Let $$I=[(1-\delta)\log n,(1+\delta)\log n]$$ and $$p_1,p_2,\dots,p_t$$ denotes all primes in $$[n/2,n]$$ and $$b_k=p_{k+1}-p_k$$. If $$A=1$$ then all $$b_k\ge(1-\delta)\log n$$. This indicates that $$\frac n2\ge\sum_{1\le k Applying Brun's sieve, I am able to recover Erdös' upper bound that $$S_1<{n\over6\log n}$$ when $$\delta$$ sufficiently small and $$n$$ sufficiently large. Mysteriously, Erdös applies this upper bound to obtain the lower estimate as follows: $$\frac n2>(1-\delta)\log n\cdot{n\over6\log n}+(1+\delta)\log n S_2,$$ and somehow he also wrote $$S_2>(\frac12-\varepsilon)n/\log n$$ so that $$\frac n2>\frac16(1-\delta)n+\frac n2(1-\varepsilon)(1+\delta)>\frac n2$$ leads to a contradiction.

I am totally confused by these steps, and I wonder whether anyone on this community can help me explain Erdös's reasoning.

Note that $$S_1+S_2=\frac{(1+o(1)) }{2}\cdot \frac{n}{\log n}$$ by the prime number theorem. Therefore writing $$S_2 =\frac{(1+o(1)) }{2}\cdot \frac{n}{\log n} - S_1$$ in your first inequality and simplifying gives
$$\frac{n}{2} \geq -2\delta S_1 \log n+ (1+\delta)\frac{1+o(1)}{2}n.$$
Using $$S_1 \log n < n/6$$ we deduce
$$\frac{n}{2} \geq n\left(\frac{1}{2}+\delta/6+o(1)\right).$$
Since $$\delta>0$$ this is a contradiction for large enough $$n$$.