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Statement

Notation (p. 251): $p_k$ is the $k$th prime.

Satz 2 (p. 251), restated. Let $p_{k_i}$, $i=1,2,\ldots$, be a subsequence of the sequence of all primes such that

$$\frac{p_{k_i}}{k_i}<\frac{p_{k_{i+1}}}{k_{i+1}}\qquad(i=1,2,\ldots).$$

Then the number of such $p_{k_i}$ with $p_{k_i}\le x$ is always $o(x/\log x)$.

The order symbol in the statement is set so that it reads as $0$ or $O$ in the print; the proof bounds every class of terms by a constant multiple of $\varepsilon x/\log x$, or by a finite number, for every $\varepsilon>0$ (pp. 253--255), and the closing remark (p. 256) calls the order of Satz 2 $o(x/\log x)$, so the statement is read with a small $o$.

Closing remark (p. 256). The paper says that the same method replaces $o(x/\log x)$ in Satz 2 by $O(x/\log^{1+\delta}x)$ for sufficiently small $\delta$, and that for example any $\delta<\frac14$ can be taken. It indicates the change: $\varepsilon$ is replaced by $(\log x)^{-\delta}$ in the proof, and the prime number theorem by the sharper relation $k=p_k/\log p_k+O(p_k/\log^2p_k)$. No further proof is written out.

Consequences stated in the paper (p. 255). The paper says that Satz 2 implies that, with the exception of at most $O(x/\log x)$ primes $p_k<x$, $p_k/k<\max_{1\le i<k}p_{k-i}/(k-i)$ (its (17)), and that with the exception of $O(x/\log x)$ such primes $p_k/k>\min_{1\le i<\infty}p_{k+i}/(k+i)$ (its (18)). As printed the exceptional sets are $O(x/\log x)$, the order of all primes up to $x$; the argument from Satz 2 gives $o(x/\log x)$ (an observation of this page).

Source. P. Erdős and K. Prachar, Sätze und Probleme über $p_k/k$, Abh. Math. Sem. Univ. Hamburg 25 (1961/1962), 251--256, doi:10.1007/BF02992930; Satz 2 on p. 251, its proof on pp. 253--255, the consequences (17) and (18) on p. 255 and the closing remark on p. 256. The edition read is identified on the source card.

Read depth. Claims checked: the statement, the consequences and the closing remark were read clause by clause on the print. The proof was read for its structure, not checked step by step. Nothing here is independently reviewed.

Proof pointer

Pp. 253--255. Fix $\varepsilon>0$ and $A=2/\varepsilon$. Fewer than $\varepsilon x/\log x$ of the chain's terms are followed by an index jump $k_{i+1}-k_i>A$. For a jump $m\le A$ the paper splits according to whether $p_{k_{i+1}}-p_{k_i}$ lies within $\varepsilon^2\log k_i$ of $m\log k_i$, falls below that window, or exceeds it. Gaps in the window are rare by the method of Satz 1; gaps below it would make $p_{k_{i+1}}/k_{i+1}\le p_{k_i}/k_i$ for large $k_i$, contradicting the chain condition by the prime number theorem; gaps above it raise $p_k/k$ by at least a fixed multiple of $\varepsilon^2\log^2x/x$, while over a range $(y/2,y]$ the ratio $p_k/k$ varies by at most $2\varepsilon^4\log x+O(1)$, which limits their number. Summing over dyadic ranges bounds these terms by $\varepsilon c_{14}x/\log x$.

Dependencies

The prime number theorem and the gap-counting estimate used for Satz 1.

Bears on

• Problem 968: context only. Satz 2 concerns a chain of indices along which $p_k/k$ increases, not the set of single steps $k$ with $p_k/k<p_{k+1}/(k+1)$ that the problem asks about, and it gives no lower bound for the density of that set.