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Claim. Let $1>a_1\ge a_2\ge\cdots>0$ and throw arcs of lengths $a_n$ independently and uniformly on a circle of unit circumference. Shepp [Sh72] proved that the arcs cover the whole circle with probability one if and only if

$$\sum_{n\ge1}\frac{e^{a_1+\cdots+a_n}}{n^2}=\infty.$$

This is the condition Problem 526 asks for, and the site records it as the solution, under the reading recorded in the problem page's Formulation. The problem's "unit circle" is read as the circle of circumference one, the setting of Dvoretzky, Kahane and Shepp (on a circle of radius one the lengths would be divided by $2\pi$, the criterion would become $\sum n^{-2}e^{(a_1+\cdots+a_n)/2\pi}=\infty$, and the site's covering example $a_n=1/n$ would fail). Shepp states the criterion for a nonincreasing sequence of lengths below $1$; Dvoretzky posed the question for any positive lengths below $1$, reading the rate of divergence on their decreasing rearrangement, and assumed monotonicity in his sufficient condition, Theorem 1 on the source card; the problem's statement writes only $a_n\ge0$, $a_n\to0$ and $\sum a_n=\infty$. Whether the arcs cover depends only on the multiset of lengths, so for lengths below $1$ in an arbitrary order the criterion is applied to their nonincreasing rearrangement, zero lengths dropped. A sequence with a length of at least $1$ is covered with probability one whatever the series does. That arc misses at most one point, and the remaining arcs, of infinite total length, cover any point placed independently of them with probability one. For example, with $a_1=1$ and $a_n=1/(2n)$ for $n\ge2$ the series converges, yet the circle is covered. The site's examples $a_n=(1+c)/n$ and $a_n=1/n$ have $a_1\ge1$ and cover for this reason. The criterion bears on their tails, since changing finitely many lengths below $1$ changes the series only by a bounded factor (a remark of this page). Taken in the given order the series condition is sufficient but not necessary, since the nonincreasing rearrangement maximizes every partial sum and a rearrangement can make the series converge while the arcs still cover (a remark of this page, not of the paper). The earlier partial results the site records, that $a_n=(1+c)/n$ (Kahane [Ka59], on his claim page) and then $a_n=1/n$ (Erdős, unpublished) cover and that $a_n=(1-c)/n$ does not, agree with the criterion applied to their tails, since $e^{a_1+\cdots+a_n}$ then grows like $n^{1+c}$, $n$ and $n^{1-c}$ (a remark of this page, not of the paper).

Depends on. No page of this wiki.

Acceptance. The site's curator, T. F. Bloom, labels the problem SOLVED and credits Shepp [Sh72] with the necessary and sufficient condition (problem page accessed), which is the reviewed evidence. The paper is refereed: L. A. Shepp, Covering the circle with random arcs, Israel J. Math. 11 (1972), no. 3, 328–345, received 1971-08-12, DOI 10.1007/BF02789327. The statement above is the paper's abstract as the publisher's record gives it; the page is dated by the issue month, September 1972. No independent proof review and no formalization are recorded.