TITLE: Certain Liouville Numbers
QUESTION [5 upvotes]: A Liouville number is a number which can be approximated very closely be a sequence of rational numbers (here is the rigorous definition I am working off of: http://en.wikipedia.org/wiki/Liouville_number).
I'm looking for an example of a Liouville number which cannot be approximated by a sequence of rational numbers with a denominators which are all a constant c multiplied by powers of some number a.
For instance, the Louiville constant ($0.110001000000000000000001$...) can be approximated by the sequence $\frac{1}{10}$, $\frac{11}{10^2}$, $\frac{110001}{10^6}$, etc, which is not what I am looking for because in each case the denominator is a power of $10$. In this case, we would say that $c=1$, $a=10$, and the denominator is always of the form $c \cdot a^n$ for some positive $n$.

REPLY [4 votes]: Consider the continued fraction
$$x = \cfrac{1}{a_1 + \cfrac{1}{a_2+ \cfrac{1}{a_3 + \ldots}}}$$
The $k$'th convergent is of the form $p_k/q_k$ where $q_k = a_k q_{k-1} + q_{k-2}$.
If $P_k$ is a prime that does not divide $q_{k-1}$, we can choose 
$a_k \mod P_k$ so that $P_k$ divides $q_k$.  Since $$\left|x - \frac{p_k}{q_k}\right| < \frac{1}{q_k q_{k+1}} < \frac{1}{a_{k+1} q_k^2}$$ taking $a_k > q_{k-1}^{k-2}$ will be enough to make $x$ a Liouville number.  Moreover,  all irreducible fractions $a/b$ such that $\left| x - \frac{a}{b} \right| < \frac{1}{2b^2}$ are convergents of $x$.  Since $q_k$ is divisible by the prime $P_k$, for any given $c$ and $d$ only finitely many of these can be of the form $c d^m$.