As Daniel Fischer noted, the length of the intervals shrinks to $0$ so the intersection is either empty or contains exactly one point.
If you choose $a_n = 0$ for each $n$, the intervals will be nested and their intersection is $\\{0\\}$. On the other hand, if you choose $a_n$ so that $\\{\frac{a_n + 1}{2^n}\\}$ is constant, i.e. $a_{n + 1} = 2a_n + 1$, and $a_0 ≥ 0$ then the intervals will be again nested but the intersection will be empty.