I´m having a hard time getting through these sum understanding exercises

This one in particular, If you have some idéa, I value some explaining more then solution so please no: Hey stupid, what the hell, just to the "inversetriangular root extraction recursive absolute value differential" method and you got it!! That just makes me even more confused when I dont understand what I am suppose to do.

The troublesome exercise is:

Look at the series of:

$\displaystyle [a_n]_{n=1}^{\infty}$ of

$\displaystyle a_{n+1}=\frac{5a_n+3}{a_n+3}$ where $\displaystyle 1\leq {n}$

with $\displaystyle a_1>0$

and $\displaystyle [b_n]_{n=1}^{\infty}$ Show that $\displaystyle b_n $is a geometric series with $\displaystyle \frac{b_n}{b_{n+1}} =$ Constant when $\displaystyle b_n$ =:

$\displaystyle b_n=\frac{a_n-3}{a_n+1}$ where $\displaystyle 1\leq{n}$

Here is my current confused work:

Well I figured lets try some different values on $\displaystyle a_1$ and se what happens, then clearly $\displaystyle a_{n+1}$ goes towards 3. Well that info could be usefull (I suppose a little sometimes). graph looks like $\displaystyle \frac{1}{x}$ graph shiften left 3 units and up 5 units.

Now lets get $\displaystyle a_n$ on one side and everything else on another and se what happens:

so:

1)$\displaystyle a_{n+1}= \frac{5a_n+3}{a_n+3}$

2)$\displaystyle a_{n+1}(a_n+3) = 5a_n+3$

3)$\displaystyle a_{n+1}*a_n+3a_{n+1} = 5a_n +3$

4)$\displaystyle 5a_n-a_{n+1}*a_n = 3a_{n+1}-3$

5)$\displaystyle a_n(5-a_{n+1}) = 3a_{n+1}-3$

6)$\displaystyle a_n= \frac{3a_{n+1}-3}{5-a_{n+1}}$

if I set $\displaystyle a_{n+1} = 3$ I get $\displaystyle a_n =3$

but if I set $\displaystyle a_n = 3$ into the $\displaystyle b_n$ formula it goes

$\displaystyle \frac{a_n-3}{a_n+1}= b_n$ = $\displaystyle \frac{3-3}{3+1} = \frac{0}{4}$

And here I am not sure what I´ve done or what I shouldn´t have done...