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 $D$.
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
$$
D([a_0;a_1,a_2,\ldots,a_n]) \le (a_1+1)(a_2+1)\cdots(a_n+1).
$$
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 $$a_{n+1} \ge \prod_{i=1}^{n}\frac{(a_i + 1)^{\mu}}{a_i^2}.$$
Best Answer
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 $k$.