Let $\{a_n\}$ be defined as follows: $a_1 = 2$, $a_{n+1}=\frac{1}{3-a_n}$, if $n \geq 1$. Does $\{a_n\}$ converge?

Let $\{a_n\}$ be defined as follows: $$a_1 = 2, a_{n+1}=\dfrac{1}{3-a_n}$$ $n \geq 1$.

Does $\{a_n\}$ converge?

Using the monotone convergence thereom, if the sequence is both bounded below and decreasing, $\forall n\in\mathbb{N}, n\geq 1$, then we can say the sequence converges.

I'm having trouble proving that the sequence is bounded below by induction.

Claim: The sequence is bounded below by $0$.

Base Case:

$n=1$

$a_1 = 2 \geq 0$, so this holds.

Induction Step

Suppose that $a_k \geq 0$, for some $k \in\mathbb{N}$. (IH)

Prove $a_k \geq 0\to a_{k+1} \geq 0$

I have some difficulty here.

$a_{k}=\dfrac{1}{3-a_{k-1}} \geq 0$

$a_{k+1}=\dfrac{1}{3-a_{k}}$, but we know $a_k \geq 0$, but we don't know how much bigger. It could be that $a_k = 4$, and then I have $a_{k+1}=\dfrac{1}{3-4}$, which is negative, and does not confirm with my claim of being bounded below by $0$. Have I gone somewhere wrong in my induction?

This is quite an overkill, but since a solution of $x=\frac{1}{3-x}$ is given by the squared golden ratio, one might wonder about Fibonacci numbers being involved. A reasonable conjecture is

$$ a_n = \frac{F_{2n-5}}{F_{2n-3}} $$ and that is straightforward to prove by induction, making the whole question trivial: the given sequence is decreasing and converging to $\frac{2}{3+\sqrt{5}}$.

Claim:

$$a_n \geq \frac{3-\sqrt{5}}{2}$$

Since $\sqrt{5} \geq 1$, we have $\sqrt{5} \geq 3-2$, $2 \geq 3-\sqrt{5}$ and hence

$$a_1=1 \geq \frac{3-\sqrt{5}}{2}$$

Suppose that we have $a_k \geq \frac{3-\sqrt{5}}{2}$,

$$3-a_k \leq 3-\frac{3-\sqrt{5}}{2}=\frac{3+\sqrt{5}}{2}$$

Assuming that you have the proof that $a_k$ is decreasing, then we know that $3-a_k >0$,

$$a_{k+1}=\frac{1}{3-a_k}\geq \frac{2}{3+\sqrt{5}}=\frac{2(3-\sqrt{5})}{9-5}=\frac{3-\sqrt{5}}{2}$$

Remark: the sequence is not bounded below by $1$. In particular, $a_3=\frac12$.

The Moebius transform $$T(z):={1\over3-z}$$ has the two fixed points $z_1={1\over2}(3+\sqrt{5})$ and $z_2={1\over2}(3-\sqrt{5})$. One computes $$T'(z)={1\over(3-z)^2}\ ,$$ so that $$T'(z_1)=\left({3+\sqrt{5}\over2}\right)^2\doteq6.85\ ,\qquad T'(z_2)=\left({3-\sqrt{5}\over2}\right)^2\doteq0.146\ .$$ The general theory of these transforms then guarantees that any recursive sequence $a_{n+1}=T(a_n)$ with $a_0\ne z_1$ converges to $z_2$, the reason being that the given $T$ is conjugate to $\hat T:\> w\mapsto \lambda w$ with $\lambda\doteq0.146$.

Try to prove that a_n is monotone decreasing and bound from below by 1/3.