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?
 A: 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 $\frac{2}{3+\sqrt{5}}$.
A: 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, $a_3=\frac12$.
A: 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 $\lambda\doteq0.146$.
A: Try to prove that a_n is monotone decreasing and bound from below by 1/3.
A: Let $\alpha$ and $\beta$ be the zero of $x^2-3x+1=0$, with $\alpha<\beta$, i.e.
$$
\alpha=\dfrac{3-\sqrt{5}}{2}\approx0.381966,\quad \beta=\dfrac{3+\sqrt{5}}{2}\approx 2.618033
$$
Claim 1: We have $\alpha<a_k<\beta$ for all $k$.
Proof:
Consider the function 
$$
f:[0,3) \to \left[\dfrac13,\infty\right),\quad f(x)=\dfrac{1}{3-x}
$$
It is clear that $f$ is continuous and increasing. Furthermore, $\alpha$ and $\beta$ are the only number satisfying $f(x)=x$. Therefore $f([\alpha,\beta])=[\alpha,\beta]$
We have
\begin{eqnarray}
a_1&=&2 \in [\alpha,\beta]\\
a_2&=&\dfrac{1}{3-2}=1\in [\alpha,\beta]\\
a_3&=&\dfrac{1}{3-1}=\dfrac12 \in [\alpha,\beta]\\
a_4&=&\dfrac{2}{5} \in [\alpha,\beta]
\end{eqnarray}
If we assume that $a_k \in [\alpha,\beta]$ for $k\le 4$, then
$$
a_{k+1}=f(a_k) \in [\alpha,\beta]
$$
Hence $a_k\in (\alpha,\beta)$ for all $k$ because $a_k\ne \alpha,\beta$ for all $k$.
Claim 2: We have $a_{k+1}<a_k$ for all $k$.
Proof:
Since $0<\alpha<\beta<3$, for all $x\in (\alpha,\beta)$ we have
$$
x-f(x)=x-\dfrac{1}{3-x}=\dfrac{x^2-3x+1}{x-3}=\dfrac{(x-\alpha)(x-\beta)}{x-3}>0,
$$
i.e.
$$
f(x)<x \quad \forall x\in (\alpha,\beta)
$$
therefore, for all $k$ we have
$$
a_{k+1}=f(a_k)<a_k \quad \forall k
$$
Claim 3: $\lim_ka_k=\alpha$.
Proof:
Since $a_k$ is decreasing and bounded below, there it is convergent, and its limit $l$ satisfies the equation $f(l)=l$, i.e. $l^2-3l+1=0$. Hence
$l\in\{\alpha,\beta\}$. It follows $l=\alpha$ because $a_k\le 2<\beta$.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
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 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
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Write $\ds{a_{n} = p_{n}/q_{n}}$ with $\ds{p_{n + 1} = q_{n}}$ and
$\ds{q_{n + 1} = 3q_{n} - p_{n}}$ which leads to $\pars{~\mbox{choose, for example,}\ p_{1} = 2\ \mbox{and}\ q_{1} = 1~}$:
\begin{align}
{p_{n + 1} \choose q_{n + 1}} & =
\pars{\begin{array}{cc}\ds{0} & \ds{1} \\ \ds{-1} & \ds{3}\end{array}}{p_{n} \choose q_{n}} =
\pars{\begin{array}{cc}\ds{0} & \ds{1} \\ \ds{-1} & \ds{3}\end{array}}^{n}{p_{1} \choose q_{1}}
\,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{\sim}\,\,\,
\lambda^{n}\,\mathbf{u}\mathbf{u}^{\mrm{T}}{2 \choose 1} =
\bracks{\lambda^{n}\mathbf{u}^{\mrm{T}}{2 \choose 1}}\mathbf{u}
\end{align}
where $\ds{\lambda}$ is the the largest magnitude eigenvalue and $\ds{\mathbf{u}}$ is the correspondent normalized eigenvector of the above matrix. It turns out that
$\ds{\mathbf{u} = {a \choose 1}/\root{a^{2} + 1}}$ where 
$\ds{a \equiv \pars{3 - \root{5}}/2}$.

Then,
\begin{align}
a_{n} & \,\,\,\stackrel{\mrm{as}\ n\ \to\ \infty}{=}\,\,\,
{\pars{1 \quad 0}\mathbf{u} \over \pars{0 \quad 1}\mathbf{u}} = {a \over 1} =
\bbx{\ds{3 - \root{5} \over 2}} 
\end{align}
