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?
 A: 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  $a_{n+1}\to 2/3$.
A: Observe that:
$$
a_{n+2} = \frac{a_{n+1} + 2}{3a_{n+1} + 2} = \frac{\frac{a_n + 2}{3a_n + 2} + 2}{3\frac{a_n  + 2}{3a_n + 2} + 2} = \frac{a_n + 2 + 6a_n + 4}{3a_n + 6 + 6a_n + 4} = \frac{7a_n + 6}{9a_n + 10}
$$
Therefore:
$$
a_{n+2} - a_n = \frac{7a_n + 6}{9a_n + 10} - a_n = \frac{-9a_n^2 - 3a_n + 6}{9a_n + 10} = \frac{-3(a_n + 1)(3a_n - 2)}{9a_n + 10}
$$
If $a_n < \frac{2}{3}$, then $\frac{2}{3} > a_{n+2} > a_n$, and if $a_n > \frac{2}{3}$ then $\frac{2}{3} < a_{n+2} < a_n$. Therefore, both subsequence $(a_{2n})$ and $(a_{2n + 1})$ are bounded and monotone, hence converge. Since the only possible limit for both subsequence is $\frac{2}{3}$, the result follows.
A: 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
$$|f(x)-f(y)|=|f'(\varepsilon)|\cdot |x-y|=\frac{4}{(3\varepsilon+2)^2}\cdot|x-y|<
\frac{4}{9}\cdot|x-y|$$
