Let the sequence $\{a_n\}$ be defined as $a_1=1$ and $a_{n+1} = \frac{6a_n+3}{a_n+4}$. Show that $a_n \lt 3$ and the sequence is increasing.

Question

Let the sequence $\{a_n\}$ be defined as $a_1=1$ and $$a_{n+1} = \frac{6a_n+3}{a_n+4}$$

Then I'm asked to show :

$1)$ $a_n \lt 3$.

$2)$ Assuming $a_n \lt 3$, show that the sequence is increasing.

For the first part I tried to use Induction. When I assumed $a_n \lt 3$ and went on to prove that this implies $a_{n+1} \lt 3$, I got stuck :
$$ a_{n+1} = \frac{6a_n+3}{a_n+4} \lt \frac{6\cdot 3+3}{a_n+4} \lt \frac{21}{a_n+4}.$$

I'm stuck here can anyone please help?

Edit : I can do the second part easily. And also I got an appropriate answer by now for the first part.

Answer 1

Observe,$$a_{n+1}=\frac{3(a_{n}+4)-(9-3a_n)}{a_n+4}=3-\frac{9-3a_n}{a_n+4}$$ If $a_n<3$, $$\dfrac{9-3a_n}{a_n+4}>0 \implies a_{n+1}<3$$