TITLE: Limit for a Recurrence Relation
QUESTION [8 upvotes]: How I can find a limit for this recursively defined sequence?
$$a_0>0, a_{n+1}=\frac{a_{n}+2}{3a_{n}+2}$$
I'm particularly interested in answers involving concepts like contractive sequences and fixed points.
Many thanks.

REPLY [5 votes]: There are two fixed points, $\frac23$ and $-1$. Let's look at the stability near each of these.
Let
$$
f(x)=\frac{x+2}{3x+2}\tag{1}
$$
Then
$$
f'(x)=-\frac43\frac1{(3x+2)^2}\tag{2}
$$
Since $f'\left(\frac23\right)=-\frac13$ and $f'(-1)=-\frac43$, $\frac23$ is a stable fixed point, $|f'(x)|\lt1$, and $-1$ is an unstable fixed point, $|f'(x)|>1$.
Let's investigate the stable fixed point. The recursive definition centered on $\frac23$ becomes
$$
\left(a_n-\tfrac23\right)=-\frac{\left(a_{n-1}-\tfrac23\right)}{3\left(a_{n-1}-\tfrac23\right)+4}\tag{3}
$$
Note that if $a_{n-1}\gt0$, then $3\left(a_{n-1}-\tfrac23\right)+4\gt2$. Thus, $(3)$ implies that
$$
\left|a_n-\tfrac23\right|\lt\tfrac12\left|a_{n-1}-\tfrac23\right|\tag{4}
$$
$(4)$ guarantees convergence.