Liouville numbers and continued fractions

Question

First, let me summarize continued fractions and Liouville numbers.

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Continued fractions.

We can represent each irrational number as a (simple) continued fraction by $$[a_0;a_1,a_2,\cdots\ ]=a_0+\frac{1}{a_1+\frac{1}{a_2+\frac{1}{\ddots}}}$$ where for natural numbers $i$ we have $a_i\in\mathbb{N}$, and we also have $a_0\in\mathbb{Z}$. Each irrational number has a unique continued fraction and each continued fraction represents a unique irrational number.

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Liouville numbers.

An irrational number $\alpha$ is a Liouville number such that, for each positive integer $n$, there exist integers $p,q$ (where $q$ is nonzero) with $$\left|\alpha-\frac pq\right|<\frac1{q^n}$$ The important thing here is that you can approximate Liouville numbers well, and the side effect is that these numbers are transcendental.

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Now if we look at the Liouville's constant, that is, $L=0.1100010\ldots$ (where the $i!$-th digit is a $1$ and the others are $0$), then we can write $$L=[0;9,1,99,1,10,9,999999999999,1,\cdots\ ]$$ The large numbers in the continued fraction make the convergents very close to the actual value, so that the number it represents is in that sense "well approximatable".

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My question now is, can we bound the numbers in the continued fraction below to be sure that the number it represents is a Liouville number?

Answer 1

Bounding the error.

The error between a continued fraction $[a_0;a_1,a_2,\ldots]$ and its truncation to the rational number $[a_0;a_1,a_2,\ldots,a_n]$ is given by $$ |[a_0;a_1,a_2,a_3,\ldots] - [a_0;a_1,a_2,\ldots,a_n]|=\left|\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots]}\right) - \left(a_0 + \frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\right|=\left|\frac{[a_1;a_2,a_3,\ldots,a_n]-[a_1;a_2,a_3,\ldots]}{[a_1;a_2,a_3,\ldots]\cdot[a_1;a_2,a_3,\ldots,a_n]}\right| \le \frac{\left|[a_1;a_2,a_3,\ldots,a_n]-[a_1;a_2,a_3,\ldots]\right|}{a_1^2}, $$ terminating with $\left|[a_0;a_1,a_2,a_3,\ldots] - [a_0;]\right|\le 1/a_1$; by iterating this recursive bound we conclude that $$ \left|[a_0;a_1,a_2,a_3,\ldots] - [a_0;a_1,a_2,\ldots,a_n]\right| \le \frac{1}{a_1^2 a_2^2 \cdots a_n^2}\cdot \frac{1}{a_{n+1}}. $$ Let $D([a_0;a_1,a_2,\ldots,a_n])$ be the denominator of the truncation $[a_0;a_1,a_2,\ldots,a_n]$ (in lowest terms). Then we have a Liouville number if for any $\mu > 0$, the inequality $$ a_{n+1} \ge \frac{D([a_0;a_1,a_2,\ldots,a_n])^\mu}{a_1^2 a_2^2 \cdots a_n^2} $$ holds for some $n$. To give a more explicit expression, we need to bound the growth of $D$.

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Bounding the denominator.

Let $D(x)$ and $N(x)$ denote the denominator and numerator of a rational number $x$ in lowest terms. Then $$ D([a_0;a_1,a_2,\ldots, a_n])=D\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\\ =D\left(\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)=N([a_1;a_2,a_3,\ldots,a_n]), $$ and $$ N([a_0;a_1,a_2,\ldots, a_n])=N\left(a_0+\frac{1}{[a_1;a_2,a_3,\ldots,a_n]}\right)\\ =N\left(a_0+\frac{D([a_1;a_2,a_3,\ldots,a_n])}{N([a_1;a_2,a_3,\ldots,a_n])}\right) = a_0 N([a_1;a_2,a_3,\ldots,a_n]) + D([a_1;a_2,a_3,\ldots,a_n]) \\ = a_0 D([a_0;a_1,a_2,\ldots, a_n]) + D([a_1;a_2,a_3,\ldots,a_n]). $$ So $$ D([a_0;a_1,a_2,\ldots,a_n]) = a_1 D([a_1;a_2,a_3,\ldots,a_n]) + D([a_2;a_3,a_4,\ldots,a_n]), $$ and the recursion terminates with $D([a_0;])=1$ and $D([a_0;a_1])=D(a_0+1/a_1)=a_1$. Since we have $D([a_0;a_1,a_2,\ldots,a_n]) \ge a_1 D([a_1;a_2,a_3,\ldots,a_n])$, we can say that $D([a_1;a_2,a_3\ldots,a_n]) \le \frac{1}{a_1}D([a_0;a_1,a_2,\ldots,a_n])$, and so $$ D([a_0;a_1,a_2,\ldots,a_n]) \le \left(a_1 +\frac{1}{a_2}\right) D([a_1;a_2,a_3,\ldots,a_n]) \le (a_1 + 1)D([a_1;a_2,a_3,\ldots,a_n]). $$ An explicit bound on the size of the denominator is therefore $$ D([a_0;a_1,a_2,\ldots,a_n]) \le (a_1+1)(a_2+1)\cdots(a_n+1). $$

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Conclusion.

We conclude the following theorem:

The continued fraction $[a_0;a_1,a_2,\ldots]$ is a Liouville number if, for any $\mu > 0$, there is some index $n$ such that $$a_{n+1} \ge \prod_{i=1}^{n}\frac{(a_i + 1)^{\mu}}{a_i^2}.$$