Let $\Delta\subset\mathbb{C}$ be the unit disk and take $f:\Delta\rightarrow \Delta$, defined by $$f(z):=\dfrac{2z+1}{z+2}.$$ It is known that $f\in Aut(\Delta)$, since we can write $$f(z)=\dfrac{z-\alpha}{1-\bar{\alpha}z},$$
with $\alpha=-\frac{1}{2}$. Hence, the behaviour of the sequence of the iterates $\{f^{\circ n}\}_n$ of $f$ is at least pseudoperiodic and $f$ cannot diverge on compact sets.
But, we can explicit the $n-$ element of the sequence as
$$f^{\circ n}(z)=\dfrac{a_n(z+1)+z}{a_n(z+1)+1},$$
where $a_n$ is defined by $a_1=1$ and $a_n=3a_{n-1}+1$, so in particular we have
$$\lim_{n\uparrow\infty}a_n=+\infty.$$
Thus what it seems to be incompatible with $f\in Aut(\Delta)$ is that
$$\lim_{n\uparrow\infty}f^{\circ n}(z)=1\qquad\forall z\in\Delta,$$
or in other words that $1$ is the Wolff point of $f$ and $\{f^{\circ n}\}_n$ diverges on any compact subset of $\Delta$.
Where am I wrong? Thank you.