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Statement

Setting (§ 2, p. 291). $\rho(x)=x-[x]$ is the fractional part; $\lg$ is the paper's notation for the logarithm.

Satz 2 (p. 293). For every $\varepsilon>0$ and every $x$ outside a set of measure zero,

$$\sum_{k=1}^{n}\rho(kx)-\frac n2=o(\lg^{1+\varepsilon}n).$$

The quantifiers are as printed: the exceptional null set is allowed to depend on $\varepsilon$. The paper remarks (p. 295) that the estimate can evidently be sharpened somewhat, but not much, as Satz 3 shows. Contrast Satz 1: no such rate holds for every irrational $x$.

Proof pointer

Pp. 293--295. With $p_k/q_k$ the convergents of $x$, $q_i\le n<q_{i+1}$ and $n=S_0q_i+R_0$, $0\le R_0<q_i$, the paper proves the inequality (1) and its consequence

$$\Bigl|\sum_{k=1}^{n}\rho(kx)-\frac n2\Bigr| <\Bigl|\sum_{k=1}^{R_0}\rho(kx)-\frac{R_0}2\Bigr|+\frac12a_{i+1}(x)+\frac32$$

(p. 294). Iterating with $R_0$ in place of $n$ ends after $O(\log n)$ steps and bounds the left side by $\frac12A_{i+1}(x)+C'\lg n$ with $C'$ absolute. Since $q_m>e^{\omega m}$ for some $\omega>0$, $i<\frac1\omega\lg n$, and the Satz of § 1 finishes the proof.

Read depth

Claims checked: the statement was read clause by clause on the page images of the print, and the proof was followed. Nothing here is independently reviewed.

Dependencies

Satz of § 1 (p. 289).

Source. A. Khintchine, Ein Satz über Kettenbrüche, mit arithmetischen Anwendungen, Math. Z. 18 (1923), 289--306; the edition read is named on the source card.

Bears on

No Erdős problem directly. The paper uses it, with the Satz of § 3, in the lattice-point lemma of § 4 (p. 301).