<s>[INST] Show that $a_n$ is decreasing 
$a_1 = 2, a_{n+1} = \frac{1}{3 - a_n}$ for $n \ge 2$. Show $a_n$ is decreasing.

First we need to show $a_n > 0$ for all $n$.
$a_2 = 1/2$ and $a_3 = 2/5$ and $a_4 = 5/13$ 
One way we can do this is by showing $3- a_n > 0$. Thus suppose it holds for $n$ then we need to show $\frac{3(3 - a_n) - 1}{3 - a_n} = \frac{8 - 3a_{n}}{3 - a_n} > 0$, which means showing $8 > 3a_{n}$, but I'm having trouble showing it.
 [/INST] First compute, $a_2=\frac{1}{3-2}=1\leq a_1<3$. Then, notice that
$$
a_{n+1}-a_{n}=\frac{1}{3-a_n}-\frac{1}{3-a_{n-1}}=\frac{a_n-a_{n-1}}{(3-a_n)(3-a_{n-1})}\leq 0
$$
which gives you the induction step for a strong induction argument for a dual hypothesis: $a_n< 3$ and $a_{n}\leq a_{n+1}$ for all $n\geq 1$. Note that if you already have $a_n<3$, then $a_{n+1}\leq a_n$ will automatically give $a_{n+1}<3$.
</s>