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.

A little playing shows that the limit "should" be $\frac{2}{3}$. So it is natural to compute $a_{n+1}-\frac{2}{3}$.
We get 
$$a_{n+1}-\frac{2}{3}=\frac{a_n+2}{3a_n+2}-\frac{2}{3}=\frac{\frac{2}{3}-a_n}{3a_n+2}.$$
 Thus 
$$\left|a_{n+1}-\frac{2}{3}\right|=\left|a_n-\frac{2}{3}\right|\frac{1}{3a_n+2}.$$
In particular, 
$$\left|a_{n+1}-\frac{2}{3}\right|\lt \frac{1}{2}\left|a_n-\frac{2}{3}\right|.$$
So with each iteration our distance from $\frac{2}{3}$ shrinks by a factor of at least $\frac{1}{2}$. It follows that $\lim_{n\to\infty}a_n=\frac{2}{3}$.
