Convergence of a recursively defined sequence of complex numbers: $3z_{n}=z_{n-1}+z_{n-2}+z_{n-1}z_{n-2}$ Question.

Let $(z_n)$ be a complex sequence such that
$$
z_0 = 0,\quad  z_1=1, \quad z_{n}=\frac13(z_{n-1}+z_{n-2}+z_{n-1}z_{n-2})\quad (n>1)\tag{1}
$$
Prove that the limit of the sequence exists.

Observations.

*

*This is a complex sequence; one cannot expect to apply the monotone convergence theorem directly.
(Added notes: Thanks to paw88789's comment below, this is actually a real sequence. The $z_n$'s are originally coefficients of a complex power series (in the linked question mentioned later) which could be complex. But I overlooked the assumption that it is immediate from the definition that they are actually all real. One could generalize this problem by replacing  $z_1$ with some non-real complex number though. But that's not the current focus. Since this sequence is actually real and bounded from below, one can directly apply the monotone convergence theorem if one can show that the sequence is "eventually" decreasing.)


*One straightforward idea is to show that the sequence is Cauchy. However, it seems difficult to directly estimate $|z_m-z_n|$ using (1).


*The recurrence in (1) is very close to linear, where one can use the characteristic function techniques. But it is not very clear how to pass from "linear" to "nonlinear".
Remarks.
This is a follow-up question to another one of mine. The question above is an essential step in one approach to establish the calculation in the linked question. I find this isolated step itself is interesting. Since I figured out a detailed proof enlighted by a sketchy comment under the linked post, I will post one proof as an answer below.
I would also like to see other approaches to the question.
 A: Updated. Thanks to Philipp's comment below, I realize that the "base step" is missing in the original argument; I find a way to fix it as below. The parameter $k$ is the original argument is also not needed.
Proof. It suffices to show that

for sufficiently large $n$, $0\le a_n\le q^n$ for some $0<q<1$ where $a_n:=|z_n|$.

By considering the function $f(x)=3x^2-x-1$ one can find that for all $n\ge 7$, $$
q^n\le 3q^2-q-1,\quad q:=0.9\tag{3}
$$
which is equivalent to
$$
\frac13(q^n+q^{n-1}+q^{2n-1})\le q^{n+1}\tag{4}
$$
Let $P(m)$ be the statement that $a_m\le q^m$. We now show by induction that $P(m)$ is true for all $m\ge 7$.
Base step. $P(7)$ is true by direct calculation by hand. (One can also write a program with a short loop to do that.)
Induction step. Suppose $P(m)$ is true for all $m$ with $7\le m\le n$ ($n>7$), we show that $P(n+1)$ is true. This follows from the induction hypothesis and (4):
$$
a_{n+1}\le \frac13(|z_n|+|z_{n-1}|+|z_n|\cdot |z_{n-1}|)\le
\frac{1}{3}(q^n+q^{n-1}+q^{2n-1})\le q^{n+1}~.
$$
Q.E.D.


[Original proof where the base step is missing.]
It suffices to show that

for sufficiently large $n$, $0\le a_n\le kq^n$ for some $k>0$ and $0<q<1$ where $a_n:=|z_n|$.

A straightforward induction shows that $0\le a_n< 1$ for $n\ge 2$. Indeed, in the induction step, by the triangle inequality:
$$
0\le a_{n+1}=|z_{n+1}|< \frac13(|z_n|+|z_{n-1}|+|z_nz_{n-1}|)<\frac13(1+1+1)=1\;.
$$
Now let us find a possible pair of $k$ and $q$ that works. Assume that $a_m\le kq^m$ for all $m\le n$. One must have
$$
a_{n+1}\le \frac13(|z_n|+|z_{n-1}|+|z_nz_{n-1}|)\le
\frac{k}{3}(q^n+q^{n-1}+kq^{2n-1})~.
$$
So if
$$
\frac{k}{3}(q^n+q^{n-1}+kq^{2n-1})\le kq^{n+1}\tag{1}
$$
then by induction $a_n\le kq^n$ for all $n$. Note that (1) is equivalent to
$$
kq^n\le 3q^2-q-1\tag{2}
$$
Consider the function $f(x)=3x^2-x-1$. Note that $f(0)=-1<0$ and $f(1)=1>0$. Hence, by the intermidiate value theorem, there exists $r\in(0,1)$ such that $f(r)=\frac12$. Let $q=r, k=1$ in (2). Then (2) is true for all sufficiently large $n$.
Consequently, we have $a_n\le r^n$ for all sufficiently large $n$.

A: Although ripples answer is pretty clever, I would like add another one where we use the monotone convergence theorem of real sequences.

The boundedness is clear. Then we do a short proof by induction to show monotonicity.
The base case/the induction starts at the index $6$ as $z_6-z_5<0$ and $z_6-z_4<0$.
Then, we assume the induction hypothesis "$z_{n}<z_{n-1}$ and $z_n<z_{n-2}$ with $n>5$".
The induction step $n\to n+1$ can be shown as follows:
$$
z_{n+1}-z_n=\frac{z_{n}+z_{n-1}+z_{n}z_{n-1}}{3}-\frac{z_{n-1}+z_{n-2}+z_{n-1}z_{n-2}}{3}\\=\frac{z_n-z_{n-2}+z_{n-1}(z_n-z_{n-2})}{3}=\frac{(1+z_{n-1})(z_n-z_{n-2})}{3}\underset{\text{hypothesis}}{\underset{\text{induction}}{<}}0
$$
and
$$
z_{n+1}-z_{n-1}<z_{n+1}-z_n\underset{\text{above}}{\underset{\text{see}}{<}}0.
$$
Hence, the real valued sequence $(z_n)$ is monotone from the $5$-th member onward. Due to monotone convergence theorem the sequence is convergent.
