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Statement

Setting (p. 1122). $\mathcal H=\{h_1,h_2,\ldots,h_{k_0}\}$ is a set of distinct nonnegative integers. For a prime $p$, $\nu_p(\mathcal H)$ is the number of distinct residue classes modulo $p$ occupied by the $h_i$, and $\mathcal H$ is admissible when $\nu_p(\mathcal H)<p$ for every prime $p$. As usual $p_n$ is the $n$-th prime.

Theorem 1 (p. 1122). If $\mathcal H$ is admissible and $k_0\ge3.5\times10^6$, then there are infinitely many positive integers $n$ for which the $k_0$-tuple

$$\{n+h_1,\ n+h_2,\ \ldots,\ n+h_{k_0}\}$$

contains at least two primes. Consequently

$$\liminf_{n\to\infty}(p_{n+1}-p_n)<7\times10^7. \tag{1.5}$$

The paper derives (1.5) on p. 1122 by taking $\mathcal H$ to consist of $k_0$ distinct primes each greater than $k_0$, which is admissible, and using $\pi(7\times10^7)-\pi(3.5\times10^6)>3.5\times10^6$. In words: there are at least $3.5\times10^6$ primes in $(3.5\times10^6,7\times10^7]$, so such an $\mathcal H$ fits inside an interval of length less than $7\times10^7$, and two primes in one translate differ by less than that.

On p. 1123 the paper says the bound in (1.5) is not optimal, that the condition $k_0\ge3.5\times10^6$ is crude and can be relaxed in certain ways, and that making the right side of (1.5) as small as possible is an open problem it does not discuss.

Source. Yitang Zhang, Bounded gaps between primes, Ann. of Math. (2) 179 (2014), no. 3, 1121--1174, DOI 10.4007/annals.2014.179.3.7, read in the journal's edition identified on the source card: Theorem 1 and (1.5) on p. 1122, the remark on optimality on p. 1123, the deduction of Theorem 1 from Theorem 2 in Sections 2, 4 and 5 (pp. 1123--1143).

Read depth. Claims checked: the setting, the statement, and the deduction of (1.5) were read clause by clause on the journal's pages. The proof was not checked step by step, and nothing here is independently reviewed.

Proof pointer

The argument follows Goldston, Pintz and Yildirim (Section 2, pp. 1123--1127). It suffices to treat $k_0=3.5\times10^6$, and the paper fixes $D=x^{1/4+\varpi}$ with $\varpi=1/1168$ and $l_0=180$ (pp. 1125--1126). With $\theta(n)=\log n$ on primes and $0$ otherwise, and the sieve weight $\lambda(n)$ of (2.11), a Goldston-Pintz-Yildirim weight restricted to divisors $d<D$ of $\prod_j(n+h_j)$ free of primes $\ge x^{\varpi}$, it compares

$$S_1=\sum_{x\le n<2x}\lambda(n)^2,\qquad S_2=\sum_{x\le n<2x}\Bigl(\sum_{i=1}^{k_0}\theta(n+h_i)\Bigr)\lambda(n)^2 .$$

If every translate $n+\mathcal H$ with $x\le n<2x$ held at most one prime, the inner sum would be below $\log 3x$ for large $x$, and $S_2\le(\log 3x)S_1$. So $S_2-(\log 3x)S_1>0$ (the paper's (2.3)) for all large $x$ gives such an $n$ in every dyadic range $[x,2x)$ with $x$ large, hence infinitely many. Section 4 (pp. 1135--1141) bounds $S_1$ above (4.20); Section 5 (pp. 1141--1143) bounds $S_2$ below (5.6), using Theorem 2 to control the error terms in the distribution of primes to smooth moduli. The two bounds give (5.7) with an explicit constant $\omega$, and the numerical check (5.8) that $\omega>0$ (p. 1143) yields (2.3).

Depends on. Theorem 2 (p. 1126) and the lemmas of Section 3 (pp. 1127--1135).

Bears on

• Problem 15: the theorem says nothing about the convergence of $\sum(-1)^nn/p_n$, the problem's question. The problem page records, as a site remark the site credits to Weisenberg, that the companion series $\sum(-1)^n/(p_{n+1}-p_n)$ diverges; (1.5) gives this, since infinitely many of its terms have absolute value greater than $1/(7\times10^7)$, so its terms do not tend to $0$.