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

Solution 1:

^{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 $n$.}