TITLE: Do the tails of the decimal expansion of pi form a dense set in [0,1]?
QUESTION [6 upvotes]: Let $a_n=10^n \cdot \pi$. Is the set of numbers $\{a_n-\lfloor a_n \rfloor : n \in \mathbb{N}\}$ dense in [0,1]?
What is the best known result near this question?
Apparently John Nash asked this on an undergraduate analysis exam (according to an anecdote told by Seymour Haber, recounted in Sylvia Nassar's biography of Nash).

REPLY [4 votes]: The comments have pretty much said all there is to be said about $\pi$. I'll just note that the fractional part of $10^n\alpha$ is known to be not just dense but uniformly distributed in $[0,1)$ for all real $\alpha$, except for a set of measure zero. There is no reason to think $\pi$ is in this exceptional set, and no expectation of a proof anytime soon that it isn't.