Find the limit of the sequence given by $a_{n+1}=3-\frac{1}{a_n}$ with $a_1=1$

Question

The answer book tells me it's $\frac{1}2(3+\sqrt 5)$ but I have no idea how to come up with this answer.

Answer 1

Using the fact that $f:x\mapsto 3-\frac{1}{x}$ is increasing on $[1,+\infty)$, you can show that $(a_n)$ is monotonic, and in fact increasing. You can also show by induction that $a_n\leq 3$ (and $a_n\geq 1$). Those two facts imply that $(a_n)$ converges. So there exists some limit $l\geq 1$.

Now you have $a_{n+1} = f(a_n)$. The left hand side converges to $l$. Besides $f$ is continuous in $l$. This implies, as $a_n\to l$, that $f(a_n)\to f(l)$. Thus $l=f(l)$. If you solve this you get two solutions. $l=\frac{3-\sqrt{5}}{2}$ or $l=\frac{3+\sqrt{5}}{2}$. The first solution does not fit here because it is $\lt 1$. So the limit is $l=\frac{3+\sqrt{5}}{2}$.

Answer 2

Hint: First prove that $a_n$ converges, say $a_n\rightarrow L$. If $\lim a_n=L$, what is $\lim a_{n+1}$? After that take limits on both sides.