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Statement

Setting (§ 2, p. 291). For irrational $x$, $\rho(x)=x-[x]$. The paper recalls Sierpiński's theorem (Krakauer Anz., math.-nat. Kl. A, Jan. 1910, p. 9) that for every irrational $x$, $\sum_{k=1}^{n}\rho(kx)-\frac n2=o(n)$, and asks whether this estimate can be sharpened.

Satz 1 (pp. 291--292). Let $\varphi(n)$ be any positive function of the integer argument $n$ with $\lim_{n\to\infty}\varphi(n)=0$. Then there is an irrational $x$ that does not satisfy

$$\sum_{k=1}^{n}\rho(kx)-\frac n2=O(n\varphi(n)).$$

So for the set of all irrationals the answer is no; the paper then turns to almost all $x$ in Satz 2.

Proof pointer

Pp. 292--293. A nested-interval construction: choose fractions $p_i/q_i$ in reduced form with increasing denominators and indices $n_1<n_2<\cdots$ so that the average of $\rho(kp_i/q_i)$ over $k\le n_i$ differs from $\frac12$ by more than $i\,\varphi(n_i)$; right continuity of $\rho$ keeps the inequality on an interval $\delta_i$ with left end $p_i/q_i$, the $\delta_i$ are nested and shrink to an irrational point, and that point satisfies the inequality for every $i$.

Read depth

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

Dependencies

None in the corpus.

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.