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 $a_n = 1$

$$a_{n+1} = 3 - \frac{1}{a_n} \text{ for n &gt; 1}$$

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 $a_2 = 2$. $a_1 &lt; a_2$ and $a_1,a_2 &gt; 0$.

$\textbf{Inductive Hypothesis:}$ $a_n &lt; a_{n+1}$ where $a_n, a_{n+1}$.

$\textbf{Inductive Step:}$ We prove that $a_n &lt; a_{n+1} \Rightarrow a_{n+1} &lt; a_{n+2}$. 

By our induction hypothesis:

$$a_n &lt; a_{n+1}$$

$$-a_n &gt; -a_{n+1}$$

$$-\frac{1}{a_n} &lt; -\frac{1}{a_{n+1}}$$ (True by our IH since $a_n, a_{n+1} &gt; 0$ and thus, $-a_n, -a_{n+1}$ share the same sign).

$$3-\frac{1}{a_n} &lt; 3-\frac{1}{a_{n+1}}$$

$$a_{n+1} &lt; a_{n+2}$$.

Note that $a_{n+1} &gt; 0$ by our IH, so $a_{n+2} &gt; 0$.

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 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 $\frac{1}{3}&lt;a_1&lt;a_2 \Rightarrow 0&lt; \color{red}{f(a_1)\leq f(a_2)} \Rightarrow \frac{1}{3}&lt;a_1&lt;\color{red}{a_2 \leq a_3}$ and by induction $\frac{1}{3}&lt; a_{n} \leq a_{n+1}$. So, the sequence is ascending and positive. 

Given $a_n &gt; \frac{1}{3} &gt; 0, \forall n$, then $\frac{1}{a_n}&gt;0 \Rightarrow a_{n+1}=3-\frac{1}{a_n} &lt; 3 \Rightarrow 0&lt;a_{n+1} &lt; 3, \forall n$. So the sequence in bounded.
In your proof of $a_{n+1}&gt;a_n$ you used that $a_n&gt;0$, but it is not proven. 

I like the following reasoning.
$$a_{n+1}-\frac{3-\sqrt5}{2}=3-\frac{3-\sqrt5}{2}-\frac{1}{a_n}=\frac{3+\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3-\sqrt5}{2}}-\frac{1}{a_n}&gt;0$$ by induction because $a_1=1&gt;\frac{3-\sqrt5}{2}.$
$$a_{n+1}-\frac{3+\sqrt5}{2}=3-\frac{3+\sqrt5}{2}-\frac{1}{a_n}=\frac{3-\sqrt{5}}{2}-\frac{1}{a_n}=\frac{1}{\frac{3+\sqrt5}{2}}-\frac{1}{a_n}&lt;0$$ by induction because $a_1=1&lt;\frac{3+\sqrt5}{2}.$

Thus, for all natural $n$ we got:
$$\frac{3-\sqrt5}{2}&lt;a_n&lt;\frac{3+\sqrt5}{2}.$$
In another hand, $$a_{n+1}-a_n=3-a_n-\frac{1}{a_n}=\frac{\left(a_n-\frac{3-\sqrt5}{2}\right)\left(\frac{3+\sqrt5}{2}-a_n\right)}{a_n}&gt;0,$$
which says that $a$ converges.