Let $x=0.121121111112\underbrace{111...1}_{4! \;times}2\underbrace{111...1}_{5! \;times}2\underbrace{111...}_{6! \;times}$. Is it a transcendental number?
As it has no periodic decimal expansion, it is irrational. I tried to prove that it is a Liouville number, i.e. $$\forall n≥1 \quad \exists p,q \in \Bbb Z, \quad 0<\left|x-\dfrac{p}{q}\right|<\dfrac{1}{q^n}$$ Maybe using rational numbers like $4/33$ or $1/9$ (their decimal expansion is "closed" to the one of $x$). I got stuck, that's why I would like to have some help.
Thank you!