Find all values $t\geq 1$ such that the sequence is increasing

Question

Given $t\geq 1$ and a recursively defined sequence $(a_n)$:

$$a_1=t,\\ a_{n+1}a_n=3a_n-2$$ find all values of $t\geq 1$ such that the sequence $(a_n)$ is increasing.

My attempt:

$$a_{n+1}a_n=3a_n-2 \Rightarrow a_n\neq 0$$ $$\Rightarrow a_{n+1}=3-\frac{2}{a_n}$$

$$a_2=3-\frac{2}{t}$$ so we need to see when

$$3-\frac{2}{t}>t$$

And we get $$t\in(1,2)$$

I don't know if this is correct...If I take for example $t=1.5$ then we have a sequence $1.5, \frac{5}{3}, 1.8,...$ which seems to be increasing but how can we be sure that this sequence doesn't start to decrease after some term?

Answer 1

$a_1=t$ $$a_{n+1}a_n=3a_n-2$$ then for increasing condition must have $$a_{n+1}=3-\frac{2}{a_n}\geq a_n$$ this leads us to inequality $$a_n^2-3a_n-2<0$$ or $$(a_n-1)(a_n-2)<0$$ that is $\color{blue}{1

Answer 2

A sequence is increasing if $a_{n+1} \geq a_n$.

Given that

$$a_{n+1} = \frac{3a_n - 2}{a_n}, a_n \neq 0$$

we get

$$a_{n+1} \geq a_n \iff \frac{3a_n - 2}{a_n} \geq a_n \iff 3a_n - 2 \geq a_n^2 \iff -a_n^2 + 3a_n - 2 \geq 0$$

Noticing that that equation has roots $1, 2$ we get

$$-(a_n - 1)(a_n - 2) \geq 0$$

And the left-hand side is a parabola upside down. Hence it is only $\geq 0$ between its roots. i.e. $-a_n^2 + 3a_n - 2 > 0$ for $a_n \in (1,2)$.

In particular, for $n=1$, we get that, if $a_1 \in (1,2)$, the sequence is strictly increasing. If $a_1 = 1, 2$ then the sequence is constant.

It also follows that if $1 < a_n < 2$ then $a_{n+1} > a_n$ and thus your example sequence (and any other with $1 < a_n < 2$) is increasing.