Expected number of long cycles in a random permutation

For a uniformly random permutation of $\{1,2,\dots,2n\}$, let $C_n$ be the number of its cycles whose length exceeds $n$. The limit $\lim_{n\to\infty}\mathbb{E}[C_n]$ equals $\ln q$ for a rational $q$. Find $q$.

$q=2$. The expected number of $k$-cycles in a random permutation of size $m$ is $\tfrac1k$ for every $k\le m$. So $\mathbb{E}[C_n]=\sum_{k=n+1}^{2n}\tfrac1k=H_{2n}-H_n\to\ln2$.