recursive succession $a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$ I'm given this recursive succession:
$a_{n+1}=\frac{a_n+2}{3a_n+2}, a_0>0$.
This is what I've done:
$L=\frac{L+2}{3L+2} \rightarrow L_1=\frac{2}{3}$ and $L_2=-1$
if $a_0 >0 $  then $a_n>0     \forall n \in N \rightarrow    
$ the succession is positive  $\forall n \in N  $and $L_2=-1$ is impossible.
if $a_n >0 $  then  $a_n+2<3a_n+2 \rightarrow a_{n+1}=\frac{a_n+2}{3a_n+2} <1$ and then all the succession is beween $0$ and $1 $ excluded.
the succession jumps back and forth the value $\frac{2}{3}$ because:
if $a_n<\frac{2}{3}$ then $a_{n+1}>\frac{2}{3}$
if $a_n>\frac{2}{3}$ then $a_{n+1}<\frac{2}{3}$
Once I arrived at this point I don't know how to conclude that the limit of the succesion is if $\frac{2}{3}$
Can someone help me to understand how to procede?

Use Banach fixed-point theorem, for $a_{n+1}=f(a_n)$ where $f(x)=\frac{x+2}{3x+2}$ and $f'(x)= \frac{-4}{(3 x + 2)^2}$. In fact, for positive $x$
$$1>\frac{x+2}{3x+2} > \frac{1}{3}$$
and, starting with $n\geq 1$, $a_n\in\left(\frac{1}{3},1\right)$. As a result, using MVT
$$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|=\frac{4}{(3\varepsilon+2)^2}\cdot|x-y|<
\frac{4}{9}\cdot|x-y|$$
