Recursive sequence depending on the parameter

Question

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?

Answer 1

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...

Answer 2

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.