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Statement

Satz 3 (p. 296). For every $x$ outside a set of measure zero (at most),

$$\Bigl|\sum_{k=1}^{n}\rho(kx)-\frac n2\Bigr|=\Omega(\lg n).$$

Here $\rho$ is the fractional part, and footnote 5 defines $\Omega$ as the negation of $O$: the left side is not $O(\lg n)$. The paper presents it as showing that Satz 2 cannot be sharpened much (p. 295).

Proof pointer

Pp. 296--297. By Bernstein's theorem, almost every $x$ has $a_n(x)=O(n\lg^2n)$ and $a_n(x)=\Omega(n\lg n)$. For such $x$, take $i$ with $a_{i+1}(x)>A(i+1)\lg(i+1)$ for an arbitrarily large $A$, an integer $s$ between $a_{i+1}(x)/3$ and $a_{i+1}(x)/2$, and $n=sq_i$; formula (1) of the proof of Satz 2 gives the lower bound (2), and the bound on $q_{i+1}$ from the growth of the partial quotients converts it into a lower bound $AL\lg n$ with $L$ independent of $n$.

Read depth

Claims checked: the statement and footnote 5 were read clause by clause on the page images of the print, and the proof was followed for structure. Nothing here is independently reviewed.

Dependencies

None in the corpus. External input: F. Bernstein's theorem (Math. Ann. 71 (1912)) on the almost-everywhere growth of partial quotients; formula (1) from the proof of Satz 2.

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.