For each $n\in\mathbb{N}$, let $I_n$ be an interval of length $1/2^{n}$. We drop each $I_n$ onto the interval $[0,1]$ uniformly at random (so that there's "wraparound" if need be). What is the expected length of the union of all the $I_n$'s?
I can see that it should be less than $3/4$, and greater than $5/8$, but can't get a precise number. In particular, if $a_k$ is the expected length of the union of the first $k$ intervals, then I believe we have $a_1 = 1/2$ and $$a_{n+1} = (1-a_n)\frac{1}{2^{n+1}} + a_n$$ But I can't figure out how to solve this recurrence and compute $\lim_{n\to\infty} a_n$.
It appears that $2^{(n+2)(n+1)/2} a_{n+1} $ is in the OEIS as sequence A114604. Unfortunately there's not much listed there about this sequence.