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?

Note that if $a_0>0$ then $a_n>0$ for all $n\geq 1$ and
$$|a_{n+1}-2/3|=\frac{|a_n-2/3|}{|3a_n+2|}\leq \frac{|a_n-2/3|}{2}.$$
Hence, 
$$|a_{n+1}-2/3|\leq \frac{|a_n-2/3|}{2}\leq 
\frac{\frac{1}{2}|a_{n-1}-2/3|}{2}=\frac{|a_{n-1}-2/3|}{2^2}\leq \dots\leq \frac{|a_0-2/3|}{2^{n+1}}.$$
Therefore, we have that $\frac{|a_0-2/3|}{2^{n+1}}\to 0$ as $n\to \infty$, and we may conclude that  $a_{n+1}\to 2/3$.
