Limit for a Recurrence Relation 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.

Use the recurrence relation 
$$ a_{n+1} - a_{n} = \frac{a_n +2 }{ 3 a_n +2} - \frac{a_{n-1} +2}{3 a_{n-1} +2 } = \frac{4(a_{n-1} -a_{n})}{(3 a_{n-1} +2)(3 a_n +2)}. $$
Since 
$$ 
3 a_n = 3 \frac{a_{n-1} +2 }{3 a_{n-1} +2} > 1 + \frac{1}{a_{n-1}+1} >1,  \quad\forall n\geq 1,
$$
it follows that
$$
|a_{n+1} - a_{n}| < \frac{4}{9} |a_{n} - a_{n-1}|.
$$
Iteration gives 
$$
|a_{n+1} - a_{n}| < \left(\frac{4}{9}\right)^n |a_{1} - a_{0}|.
$$
The series $\sum_{n=1}^\infty (a_{n+1}-a_n)$, of positive terms, is dominated by the convergent series $|a_1-a_0| \sum_{n=1}^\infty (4/9)^n$ and so converges. We have  $\sum_{n=1}^\infty (a_{n+1}-a_n)= \lim_{n\to \infty} a_n - a_1$ which shows that the limit exists.
Then, to find to fixed points we can pass to the limit in the recurrence relation
$$
a_{\infty}= \frac{a_{\infty}+2}{3 a_{\infty} +2},
$$
which leads to $a_{\infty}= 2/3.$
