Recursive sequence depending on the parameter 
For the given parameter $\mathbb R\ni t\geq 1$, the sequence is
  defined recursively: $$a_1=t,\;\;a_{n+1}a_n=3a_n-2$$ $(a)$ Let $t=4$.
  Prove the sequence $(a_n)$ converges and find its limit.
$(b)$ Which parameters $t\geq 1$ is the sequence $(a_n)$ increasing
  for?

My attempt:

Bolzano-Weierstrass:A sequence converges if it is monotonous and
   bounded

$$a_{n+1}a_n=3a_n-2\implies a_{n+1}=3-\frac{2}{a_n}$$
$(a)$ 
First few terms: $a_1=4,a_2=\frac{5}{2},a_3=\frac{11}{5}$
Assumption: the sequence is decreasing
Proof by induction:
the basis (n=1) is trivial: $\frac{5}{2}<4$
Assumption: $a_n<a_{n-1},\;\forall n\in\mathbb N$
Step: $$a_n<a_{n-1}\implies\frac{1}{a_n}\geq\frac{1}{a_{n-1}}\Bigg/\cdot(-2)$$
$$\iff-\frac{2}{a_n}\leq-\frac{2}{a_{n-1}}\iff \underbrace{3-\frac{2}{a_n}}_{a_{n+1}}\leq\underbrace{3-\frac{2}{a_{n-1}}}_{a_n}$$
The limit: $$L=3-\frac{2}{L}\implies L^2-3L+2=0$$
I take into account only $2$ because the parabola is convex and $$a_n\to L^-.$$
Then I have to prove: $a_n\geq 2\;\forall n\in\mathbb N$ after the formal computing: $a_{n+1}\geq 3-\frac{2}{2}=2$
$\underset{\implies}{\text{Bolzano-Weierstrass theorem}}(a_n)\to 2$
$(b)$ Since the sequence doesn't have to be convergent, only increasing:
$$a_2=3-\frac{2}{t}\geq t\implies t\in[1,2]$$
Then, it should follow inductively,analogously to $(a)$, this time it is increasing.
Is this correct?
 A: Let us find the general form for the sequence given by the recursion
$$
a_{n+1}=
\underbrace{
\begin{bmatrix}3&-2\\ 1&0
\end{bmatrix}
}_{}\cdot a_n\ ,
$$
where we use the Möbius action of matrices $2\times 2$ on scalars, given in general by 
$$
\begin{bmatrix}a&b\\ c&d
\end{bmatrix}\cdot x
:=
\frac{ax+b}{cx+d}\ ,
$$
see also Möbius transformation, wiki page.
The special matrix $A$ used in the problem can be diagonalized,
$$
A=
\underbrace{
\begin{bmatrix}1&1\\ 1/2&1
\end{bmatrix}}_{T}
\underbrace{
\begin{bmatrix}2&\\ &1
\end{bmatrix}}_{D}
\underbrace{
\begin{bmatrix}2&-2\\-1&2
\end{bmatrix}}_{T^{-1}}
$$ 
and because $A^n=TD^nT^{-1}$ we get the general form for $a_n=A^n\cdot\begin{bmatrix}4\\1\end{bmatrix}$, and then passing to the element in the projective space, by taking the quotient, it is:
$$
a_n=\frac{6\cdot 2^n-2}{3\cdot 2^n-2}\ .
$$
It converges to $6/3=2$.
For the part (b) a similar study can be started, the general term being
$$
a_n(t)=
TD^nT^{-1}
\begin{bmatrix}t\\ 1
\end{bmatrix}_{\Bbb P^1}
=
\begin{bmatrix}
2\cdot 2^n-1 & -2\cdot 2^n+2\\
2^n-1 & -2\cdot 2^n+2
\end{bmatrix}
\begin{bmatrix}
t\\ 1
\end{bmatrix}
\text{ considered in }
{\Bbb P^1} 
\ .
$$
I am stopping here...
A: If you have already proven $a_n > 2, \forall n$ , then
Let $b_n=a_n-2$, $b_n>0$. $a_{n+1} = 3 - \frac{2}{a_n} \Rightarrow b_{n+1} + 2=3-\frac{2}{b_n+2} \Rightarrow b_{n+1} = \frac{b_n}{2+b_n} < \frac{b_n}{2}$.
Therefore as $n\to \infty, b_n \to 0, a_n \to 2.\blacksquare$
The next is pure hindsight based on dan_fulea's solution but I believe it can be useful when the two fixed points are distinct.
$a_{n+1} - 1 = 2-\frac{2}{a_n} = \frac{2(a_n - 1)}{a_n}$
$a_{n+1} - 2 = 1-\frac{2}{a_n} = \frac{a_n - 2}{a_n}$
Therefore $\frac{a_{n+1}-1}{a_{n+1}-2} = 2\cdot \frac{a_n-1}{a_n-2}$
$\frac{a_n-1}{a_n-2}$ is a geometric sequence with initial value $\frac{3}{2}$,
so $\frac{a_n-1}{a_n-2} = 2^{n-1} \frac{3}{2} = 1+\frac{1}{a_n-2} \Rightarrow a_n = 2+ \frac{1}{2^{n-1}\frac{3}{2}-1} = \frac{6\cdot 2^{n-1}-2}{3\cdot 2^{n-1}-2}.\blacksquare$
In general if there are two distinct fixed points $r$ and $s$ then the ratio $\frac{a_n-r}{a_n-s}$ is a geometric sequence.
