Expected summation of dropped intervals?

Question

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 is "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 not 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 not figure out how to solve this recurrence and compute $\lim_{n\to\infty} a_n$. Does anyone have any insight they could lend?

Answer 1

Each point in $[0,1)$ has probability $p=\prod_{i=1}^\infty (1-2^{-i})$ of being missed by all the intervals, so the expected measure of what is not missed is $1-p$. This is approximately 0.7112119; I forget if there is a closed form.

Answer 2

This is a linear first order recurrence, so we can certainly solve it. Let me denote by $A_n$ the solution to the homogeneous equation $A_{n+1}=(1-2^{-n-1})A_n$ with the initial value $A_1=1/2$ (for convenience), so $$ A_n = \prod_{k=1}^n \left( 1 - 2^{-k} \right) . $$ Write (= variation of constants) $a_n=C_n A_n$; then $C_n$ solves $C_{n+1}-C_n =2^{-n-1}/A_{n+1}$, $C_1=0$. This gives, after manipulating, $$ \lim a_n = \frac{1}{2} \prod_{k\ge 2} \left( 1 - 2^{-k} \right) + \frac{1}{4} \prod_{k\ge 3} \left( 1 - 2^{-k} \right) + \frac{1}{8} \prod_{k\ge 4} \left( 1 - 2^{-k} \right) + \frac{1}{16} \prod_{k\ge 5} \left( 1 - 2^{-k} \right) + \ldots $$ and this formula makes me somewhat skeptical if there can be a very explicit answer. Of course, I could be missing something. (Since everything converges exponentially fast, we should be able to extract bounds on $\lim a_n$ without much trouble.)