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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 $$ \frac n2\ge\sum_{1\le k<t}b_k=(1-\delta)\log n\underbrace{\sum_{b_k\in I}1}_{S_1}+(1+\delta)\log n\underbrace{\sum_{b_k\notin I}1}_{S_2}, $$ 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.

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1 Answer 1

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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 $n$.

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