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 $n$.

  • $\begingroup$ It is not clear to me what you mean. Do you mean you are looking for a Liouville number which does not satisfy the condition in the Wikipedia article for rational numbers p/q such that q is always c times a power of a? $\endgroup$ – Qiaochu Yuan Aug 17 '10 at 5:54
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    $\begingroup$ How about $1 + 1/2^{2!} + 1/3^{3!} + 1/4^{4!} + \ldots$? $\endgroup$ – anon Aug 17 '10 at 6:31
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    $\begingroup$ Personally I think anon's comment should be an answer. $\endgroup$ – Mark Beadles Dec 31 '11 at 22:17
  • $\begingroup$ Except that proving it's Liouville might not be so simple. However, modifications can be made... $\endgroup$ – Robert Israel Jan 1 '12 at 3:46
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    $\begingroup$ Proving that anon's series is a Liouville number is not hard: the tail starting at the $(n+1)$st term is at most $2/(n+1)^{(n+1)!}$, while the denominator of the sum of the first $n$ terms is at most $n^{2n!}$. Therefore the truncation at the $n$th term gives a rational approximation of the form $|c - p/q| < 1/q^{n+1}$ (where $c$ is the sum of the infinite series). $\endgroup$ – Greg Martin Jan 1 '12 at 5:26

I wrote a paper several years back, Absolutely Abnormal Numbers, in which there is a construction of a Liouville number where the $k$th rational approximation has denominator that is a power of $k$.


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 $c d^m$.


I believe all known Liouville numbers are constructive, and pretty much by the same general idea. Some known Liouville numbers (where $c$ is Liousville's original constant):

  • $3.140001000000000000000005...$ where: each digit is $0$ where $c$ has a zero; each digit is the next digit of $\pi$ where $c$ has a $1$
  • $\xi=c + \frac{p}{q}\ |\ p,q\in\mathbb{Z} $ since the sum of a rational and a Liouville number is a Liouville number
  • $\xi=\sum_{n=1}^\infty\frac{1}{2^{in!}}$ for $i\geqslant1$
  • Similarly $\xi=[0;10,10!,(10!)!,((10!)!)!,...]$
  • $T=\sum_{n=1}^\infty\frac{1}{T_n}=\frac{1}{2}+\frac{1}{4^2}+\frac{1}{8^{4^2}}+\cdot\cdot\cdot$ where $T_n$ are power towers

Informally, the general idea is that the terms in the continued fraction of a Liouville number increase "fast enough".


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