Prove that if $2 < y_0 < 3$ then the sequence is strictly increasing but that $y_n<3$ for all $n\,$

algebra-precalculussequences-and-series

A sequence $$t_0$$, $$t_1$$, $$t_2$$, $$…$$ is said to be strictly
increasing
if $$t_{n+1} > t_n$$ for all $$n\ge{0}\,$$.

The terms of the sequence $$y_0\,$$,$$y_1\,$$, $$y_2\,$$, $$\ldots$$
satisfy $$y_{n+1}= 5-\frac 6 {y_n}$$ for $$n\ge{0}\,$$. Prove that
if $$2 < y_0 < 3$$ then the sequence is strictly increasing but that
$$y_n<3$$ for all $$n\,$$.

Hello. I am not very good at these types of questions and require some assistance.

Workings: I have started off with $$y_{n+1} – y_n > 0 \\ \implies 5-\frac{6}{y_n} – y_n>0 \\\implies \frac{-y_n^2+5y_n-6}{y_n}>0$$

So, from this, we see that when $$2 the sequence is increasing. Additionally, it's increasing when $$y_n<0$$. However, I am not sure why the question didn't include that part? Anyway, I don't know how to make any progress from $$2.

A different solution than the one provided by Atila Correia, following the same thought process you initially did, is the following:

The problem states that

$$2 < y_0 < 3 \tag{1}$$

It's also known that

$$y_{n+1} = 5 - \frac{6}{y_n} \tag{2}$$

Starting from the fact that the sequence is strictly increasing (what we want to prove) we have:

$$$$y_{n+1} - y_{n} = (5 - \frac{6}{y_n}) - y_n = \frac{-y_{n}^2 + 5y_n - 6}{y_n} > 0 \tag{3}$$$$

Thus,

$$\begin{cases} -y_n^2 + 5y_n - 6 > 0\\ y_n > 0 \tag{4} \end{cases}$$

or

$$\begin{cases} -y_n^2 + 5y_n - 6 < 0\\ y_n < 0 \tag{5} \end{cases}$$

Based on $$(1)$$, $$y_n$$ can't be negative, then $$(5)$$ is impossible. From $$(4)$$ we have

$$$$-y_n^2 + 5y_n - 6 > 0 \Rightarrow \begin{cases} y_n > 2\\ y_n < 3 \end{cases} \tag{6}$$$$

Thus $$y_n < 3 \ \forall\ n \geq 0$$, if the sequence is strictly increasing.

For the cases that the sequence is not strictly increasing we should have:

$$\begin{cases} -y_n^2 + 5y_n - 6 > 0\\ y_n < 0 \tag{7} \end{cases}$$

or

$$\begin{cases} -y_n^2 + 5y_n - 6 < 0\\ y_n > 0 \tag{8} \end{cases}$$

From $$(1)$$ we have that $$y_n$$ can't be negative for all $$n$$, so $$(7)$$ doesn't hold.

Let's call $$f$$ the $$-y_n^2 + 5y_n - 6$$ curve. For $$y_n > 0$$, $$f$$ behaves as follows:

$$\begin{cases} 0 < y_n < 2 & \implies f < 0\\ 2 < y_n < 3 & \implies f > 0\\ 3 < y_n & \implies f < 0 \tag{9} \end{cases}$$

So $$(8)$$ only holds if $$y_n$$ is not in the $$(2, 3)$$ interval. That doesn't satisfy $$(1)$$, meaning $$(8)$$ also doesn't hold. This means the sequence is necessarily strictly increasing.