A sequence $t_0$, $t_1$, $t_2$, $…$ is said to be

strictlyif $t_{n+1} > t_n$ for all $n\ge{0}\,$.

increasingThe 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 $2<y_n<3$ 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 $2<y_n<3$.

## Best Answer

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:

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

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

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

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

For the cases that the sequence is

notstrictly 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

notin the $(2, 3)$ interval. That doesn't satisfy $(1)$, meaning $(8)$ also doesn't hold. This means the sequence isnecessarilystrictly increasing.