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?
Best Answer
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...