 Longest increasing cyclic run of a random permutation

Arrange a uniform random permutation of $n$ distinct ordered elements around a circle, and let $L$ be the greatest number of consecutive entries that appear in increasing order. Put $k_0=\min \{k:k!\ge n\}$. Then $\mathbb{P}(k_0-2\le L\le k_0)\to 1$ as $n\to \infty$.

Let $X_k$ count increasing consecutive blocks of length $k$. Each of the $n$ starting positions succeeds with probability $1/k!$, so $\mathbb{E}{X}_k=n/k!$. For $k\le n/2$, disjoint blocks have independent relative orders; blocks overlapping with a start displacement $s<k$ are both increasing precisely when their length-$k+s$ union is increasing. ThusConsequently Markov's inequality gives $\mathbb{P}(L\ge k_0+1)\le 1/(k_0+1)$, while Chebyshev's inequality gives $\mathbb{P}(L<k_0-2)\le 2/(k_0-1)$ for sufficiently large $n$. The latter uses $n/(k_0-2)!>k_0-1$. Both bounds vanish. These are consecutive cyclic runs; allowing arbitrary skipped entries is a different permutation statistic.