Determine the whether the sequence $a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$ is convergent or divergent. Consider the recursively defined sequence $a_n = 1$
$$a_{n+1} = 3 - \frac{1}{a_n} \text{ for n > 1}$$
Is the sequence convergent?
This is my attempt:
First, we prove that the sequence is positive, and monotonically increasing, using induction.
$\textbf{Base Case:}$ $a_1 = 1$ and $a_2 = 2$. $a_1 < a_2$ and $a_1,a_2 > 0$.
$\textbf{Inductive Hypothesis:}$ $a_n < a_{n+1}$ where $a_n, a_{n+1}$.
$\textbf{Inductive Step:}$ We prove that $a_n < a_{n+1} \Rightarrow a_{n+1} < a_{n+2}$. 
By our induction hypothesis:
$$a_n < a_{n+1}$$
$$-a_n > -a_{n+1}$$
$$-\frac{1}{a_n} < -\frac{1}{a_{n+1}}$$ (True by our IH since $a_n, a_{n+1} > 0$ and thus, $-a_n, -a_{n+1}$ share the same sign).
$$3-\frac{1}{a_n} < 3-\frac{1}{a_{n+1}}$$
$$a_{n+1} < a_{n+2}$$.
Note that $a_{n+1} > 0$ by our IH, so $a_{n+2} > 0$.
Now we prove that the sequence is bounded.
Observe that $a_n$ is monotonically increasing, which means that $ - \frac{1}{a_n}$ is monotonically increasing as well and it is upped bounded by $0$. Thus, $3-\frac{1}{a_n}$ is upper bounded by $3$. 
We have a sequence that is monotone and bounded. Hence, by the Monotone Convergence Theorem, This sequence converges. 
I was wondering if this method is correct.
 A: A shorter approach, look at the function $f(x)=3-\frac{1}{x}$ because this function "generates" the sequence, i.e. $f(a_n)=a_{n+1}$. This function is ascending, because $f'(x)=\frac{1}{x^2}$ and $\frac{1}{3}<a_1<a_2 \Rightarrow 0< \color{red}{f(a_1)\leq f(a_2)} \Rightarrow \frac{1}{3}<a_1<\color{red}{a_2 \leq a_3}$ and by induction $\frac{1}{3}< a_{n} \leq a_{n+1}$. So, the sequence is ascending and positive. 
Given $a_n > \frac{1}{3} > 0, \forall n$, then $\frac{1}{a_n}>0 \Rightarrow a_{n+1}=3-\frac{1}{a_n} < 3 \Rightarrow 0<a_{n+1} < 3, \forall n$. So the sequence in bounded.
A: In your proof of $a_{n+1}>a_n$ you used that $a_n>0$, but it is not proven. 
I like the following reasoning.
$$a_{n+1}-\frac{3-\sqrt5}{2}=3-\frac{3-\sqrt5}{2}-\frac{1}{a_n}=\frac{3+\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3-\sqrt5}{2}}-\frac{1}{a_n}>0$$ by induction because $a_1=1>\frac{3-\sqrt5}{2}.$
$$a_{n+1}-\frac{3+\sqrt5}{2}=3-\frac{3+\sqrt5}{2}-\frac{1}{a_n}=\frac{3-\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3+\sqrt5}{2}}-\frac{1}{a_n}<0$$ by induction because $a_1=1<\frac{3+\sqrt5}{2}.$
Thus, for all natural $n$ we got:
$$\frac{3-\sqrt5}{2}<a_n<\frac{3+\sqrt5}{2}.$$
In another hand, $$a_{n+1}-a_n=3-a_n-\frac{1}{a_n}=\frac{\left(a_n-\frac{3-\sqrt5}{2}\right)\left(\frac{3+\sqrt5}{2}-a_n\right)}{a_n}>0,$$
which says that $a$ converges.
