Let ${(a_n)}$ be the sequence defined by $a_1 = 1$ and $a_{n+1} = 3 − 1/a_n$ for all $n ≥ 1$. Let ${(a_n)}$ be the sequence defined by $a_1 = 1$ and $a_{n+1} = 3 − 1/a_n$ for all $n ≥ 1$. Prove that ${a_n}$ converges and find its limit.

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*I believe I am supposed to prove that $a_n$ is a monotone sequence first and that it is bounded. Then show that in converges to let's say L. I am not sure if  this is the way I am supposed to go. Aside from this I am stuck, any help would me much appreciated.

 A: When nothing better comes to mind, it doesn’t hurt to gather a little data to get a clearer picture of what’s going on. Sometimes the data aren’t terribly helpful, sometimes they suggest a standard approach — and sometimes they can suggest a very different approach peculiar to the particular problem at hand.
$$\begin{array}{rcc}
n:&1&2&3&4&5&6\\
a_n:&\frac11&\frac21&\frac52&\frac{13}5&\frac{34}{13}&\frac{89}{34}
\end{array}$$
At the very least one can see that so far, at least, the sequence is increasing, so that it would be reasonable to try to prove this, most likely by induction on $n$. But in this case many will immediately recognize those numerators and denominators as Fibonacci numbers; specifically, the data so far suggest that
$$a_n=\frac{F_{2n-1}}{F_{2n-3}}\tag{1}$$
for $n>1$. In fact if you use its recurrence to project the Fibonacci sequence backwards, you find that $F_{-1}=1$, so that $(1)$ holds for $1\le n\le 6$. It’s easy to verify this conjecture if one has played a bit with the Fibonacci numbers. If $(1)$ holds for some $n$, then
$$\begin{align*}
a_{n+1}&=3-\frac1{a_n}\\
&=3-\frac1{\frac{F_{2n-1}}{F_{2n-3}}}\\
&=3-\frac{F_{2n-3}}{F_{2n-1}}\\
&=\frac{3F_{2n-1}-F_{2n-3}}{F_{2n-1}}\\
&=\frac{2F_{2n-1}+(F_{2n-1}-F_{2n-3})}{F_{2n-1}}\\
&=\frac{F_{2n-1}+(F_{2n-1}+F_{2n-2})}{F_{2n-1}}\\
&=\frac{F_{2n-1}+F_{2n}}{F_{2n-1}}\\
&=\frac{F_{2n+1}}{F_{2n-1}}\\
&=\frac{F_{2(n+1)-1}}{F_{2(n+1)-3}}\,,
\end{align*}$$
so $(1)$ holds for $n+1$ as well. And since $(1)$ holds for $n=1$, it must hold for all positive integers $n$. Thus,
$$a_n=\frac{F_{2n-1}}{F_{2n-3}}=\frac{F_{2n-1}}{F_{2n-2}}\cdot\frac{F_{2n-2}}{F_{2n-3}}\,,$$
and if one is familiar with the well-known fact that $\lim_\limits{n\to\infty}\frac{F_{n+1}}{F_n}=\varphi=\frac{1+\sqrt5}2$, it’s immediately clear that
$$\lim_{n\to\infty}a_n=\varphi^2=\frac{3+\sqrt5}2\,.$$
Obviously this approach is not a general technique and requires a bit of background knowledge, and overall it’s no easier than the standard approach suggested by Steve Morris in his answer, but it is an interesting surprise, and I do want to emphasize the suggestion in the first paragraph.
A: Use induction to prove that sequence is increasing and it is trivial to prove that each term in the sequence is never greater than three.
A: $$A_{n+1}=3-\frac{1}{A_n}$$
Let $A_{n}=\frac{B_{n+1}}{B_n}$, then
$$\frac{B_{n+2}}{B_{n+1}}-3+\frac{B_n}{B_{n+1}}=0 \implies  B_{n+2}-3B_{n+1}+B_N=0.$$
Let $B_n= x^n$, then $x^2-3x+1=0 \implies x_{1,2}=\frac{3\pm\sqrt{5}}{2}=a,b.$
So $$B_n= p \left(\frac{3+\sqrt{5}}{2}\right)^n+ q \left ( \frac{3-\sqrt{5}}{2}\right)^n$$
Then $$A_n=\frac{pa^{n+1}+q b^{n_|+1}}{ a^n+ qb ^n}= \frac{a^{n+1}+r b^{n_|+1}}{ a^n+ rb ^n}, q=b/a~~~~(1)$$
Using $A_1=1$, we get $$r=\frac{a^2-a}{b-b^2}=-\frac{2a+1}{2b+1}=-\frac{21+8\sqrt{5}}{11}.$$
Finally, from (1), $$\lim_{n \to \infty}A_n=a=\frac{3+\sqrt{5}}{2}$$
