Is the following sequence equidistributed mod$1$ : $x_n=\{10^n\cdot a\}_{n\in\mathbb N}, a\in\mathbb R\setminus\mathbb Q ?$

Question

Is the following sequence equidistributed mod$1$? $\big($ or equidistributed in $[0,1]\big)$ : $$x_n=\{10^n\cdot a\}_{n\in\mathbb N}, a\in\mathbb R\setminus\mathbb Q.$$

I'm familiar with the Weyl's criterion, but i'm not sure if it works here...

Answer 1

No. For example let $a=\sum_{k=1}^\infty 10^{-k!}\in\mathbb R \setminus \overline{\mathbb Q}$ be the Liouville's constant, then $x_n<0.2$ for any $n$.

Answer 2

@just-a-user gave a great short answer, but in case someone doesn’t see why the given $a$ works, here’s a longer (but basically equivalent) answer.

If $\ a\!\!\!\mod\!\!1\ $ has decimal expansion $0.d_1 d_2 d_3 d_4\!\dots$, the sequence $x_n$ is $$x_1=0.d_1 d_2 d_3 d_4\!\dots\\ x_2=0.d_2 d_3 d_4d_5\!\dots\\ x_3=0.d_3 d_4d_5d_6\!\dots\\ x_4=0.d_4d_5d_6d_7\!\dots\\ \dots .$$

If the digits $d_i$ are all, say, $0$ and $1$, then these $x_n$ are all less than $0.2$.

So let $a$ be your favorite irrational number with only $0$s and $1$s in its decimal expansion, and the sequence of $x_n\!\!\!\mod\!\!1$ will be far from equidistributed. Liouville’s constant $a=0.110001000000000000000001000\dots$ is one choice for $a$, but there are many others!