> 
> 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 $2<y\_n<3$ 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 $2<y\_n<3$.