Exploration of recurrence relation $a_{n+1}=2-\frac{1}{a_{n}}$ How can one find the limit of the recurrence equation
$a_{n+1}=2-\frac{1}{a_{n}},  \;\;n\geq 1.$
I am to prove that the sequence {$a_{n}$} converges and to find the limit.
Here is what I've done.
If $a_{1}>1$, it implies that $a_{2}=2-\frac{1}{a_{1}}\geq \frac{1}{2}.$ Assuming that the statement holds for $n=k$ then
$a_{k+2}=2-\frac{1}{a_{k+1}}\leq 2-\frac{1}{a_{k}}=a_{k+1}.$ This shows that $a_{k+2}\leq a_{k+1}$ and ${a_{n}}$ is monotone decreasing. Thus, convergent. I'd like to know if I'm correct and also find the limit of ${a_{n}}$. 
 A: Assuming, $a_n >0$ you can proceed in the following way:
$a_{n+1}=2-\frac{1}{a_{n}}$ $\Rightarrow$ $a_{n+1}-a_{n}=2-a_{n}-\frac{1}{a_{n}}=2-\left(a_n+\frac{1}{a_{n}}\right)\le0$ because $a_{n}+\frac{1}{a_{n}} \geq 2$ (here is important that $a_{n}\ge 0$ such that to be able to apply inequality mean). So, we proved that $a_{n+1}-a_{n}<0$, which means that the $(a_{n})_{n\ge1}$ is decreasing. Now we have to prove that $a_{n}$ has a superior and a inferior margin to the left and to the right.
So:
If you have $a_{n}>0$ (here, again is important to have $a_{n}>0$) and from the statement we have that $a_{n+1}=2-\frac{1}{a_{n}}\le 2$. So, $0 \le a_{n} \le 2$.
Having $a_{n+1}<a_{n}$ and $0<a_{n}<2$ we have proved that $a_{n}$ is converged. So, we can say that $$\lim_{n \to \infty} a_{n}=l \Rightarrow l=2-\frac{1}{l} \Rightarrow l^2-2l+1=0 \Rightarrow (l-1)^2=0 \Rightarrow l=1 \in (0,2).$$ 
A: 
A point that every decent proof should deal with is first and foremost whether the whole sequence $(a_n)_{n\geqslant1}$ exists, for some given initial condition $a_1$, or if it happens that some $a_n=0$ for some $n$ (and then, for every $k\geqslant n+1$, $a_k$ is undefined).

First note that, if $a_n$ exists, if $a_n\ne1$ and if $a_n\ne0$, then $a_{n+1}$ exists and $a_{n+1}\ne1$ hence $$\frac1{a_{n+1}-1}=\frac{a_n}{a_n-1}=\frac1{a_n-1}+1$$ As a consequence, if $a_1=0$, the sequence $(a_n)_{n\geqslant1}$ is undefined. If $a_1=1$ then $a_n=1$ for every $n$ hence the convergence of the sequence $(a_n)_{n\geqslant1}$ and its limit are clear. And if $a_1\ne0$ and $a_1\ne1$, then, as long as $a_k\ne0$ and $a_k\ne1$ for every $1\leqslant k\leqslant n$, one has $$\frac1{a_n-1}=\frac1{a_1-1}+n-1$$
This proves that the cases when the sequence $(a_n)_{n\geqslant1}$ is not completely defined are when $$a_1=1-\frac1k$$ for some $k\geqslant1$ (and then $a_i$ exists for every $1\leqslant i\leqslant k$, but not $a_{k+1}$). 

For every other initial condition $a_1$, hence in particular for every $a_1\geqslant1$, the sequence $(a_n)_{n\geqslant1}$ is fully defined and, for every $n\geqslant1$, $$a_n=1+\frac{a_1-1}{1+(n-1)(a_1-1)}=\frac{n(a_1-1)+1}{(n-1)(a_1-1)+1}$$

Using this, the convergence of the sequence $(a_n)_{n\geqslant1}$ and its limit in this case should be clear.
