Finding the limit of a recursive sequence. 
Let $a_n \in \Bbb{R}$ such that: $$ a_{n+1}=1+\frac{2}{a_n}  \text{  and   } a_1=1$$
Prove that $\lim_{n \to +\infty}a_n=2$

We have that $a_n \geq 1,\forall n \in \Bbb{N}$ thus $$\limsup_na_n \geq \liminf_na_n \geq 1$$
Let $\liminf_na_n=l$
Then we have that $l^2-l-2=0$ thus $l=2,-1$
But $a_n \geq 1$ thus $l=2$.
Applying the same argument we prove that $\limsup_na_n=2$
Also we can also derive a contradiction if we assume that the limit superior and inferior are infinite.
Is my attempt an efficient way to prove this statement?
Also can someone provide additional solutions?
Thank you in advance.
 A: Banach's Fixed Point theorem also provides the limit.
It states that the iteration $a_{n+1} = f(a_n)$ will converge to  $a^* = f(a^*)$ if $|f'(a)| <1$ during  the iteration. 
Now $|f'(a)| = 2/a^2$ which in general will not be less than one. Taking two iteration steps however gives 
$$
a_{n+2} = g(a_n) = 3 - \frac{4}{a_n+2}
$$
and $|g'(a)| = \frac{4}{(a+2)^2}$ which is less than one for all $a_n >0$, which will always be the case over the iteration.  
A: Here we have where $a_{n+1}=f(a_n)$ where
$$f(x)=1+\frac2x=\frac{x+2}{x}.$$
Maps of the form $x\mapsto (ax+b)/(cx+d)$ are fractional linear transformations and can be represented by matrices. If we let $A=\pmatrix{a&b\\c&d}$ and define
$$A\bullet x=\frac{ax+b}{cx+d}$$
then $A\bullet (B\bullet x)=AB\bullet x$.
Here take $A=\pmatrix{1&2\\1&0}$. Then $a_{n+1}=A^n\bullet1$. We can diagonalise $A$:
$$A=\pmatrix{1&2\\-1&1}\pmatrix{-1&0\\0&2}\pmatrix{1&2\\-1&1}^{-1}$$
and so
$$A^{n-1}=\pmatrix{1&2\\-1&1}\pmatrix{(-1)^n&0\\0&2^n}
\pmatrix{1&2\\-1&1}^{-1}.$$
Now it is straightforward to calculate Dr Graubner's formula.
A: an explicit formula is given by $$a_n=-\frac{2+\left(\frac{-1}{2}\right)^n}{-1+\left(\frac{-1}{2}\right)^n}$$ so $$\lim_{n\to \infty}a_n=2$$
A: Hint:


*

*$a_n \ge 2 \implies a_{n+2} \ge 2, \ a_{n+2} \le a_n$

*$a_n \le 2 \implies a_{n+2} \le 2, \ a_{n+2} \ge a_n$

*$ 1 = a_1 \le a_3 \le a_5 \le \cdots \le 2 \le \cdots \le a_6 \le a_4 \le a_2 =3 $
A: $a_{n+2}-a_n=\frac{4(a_n-a_{n-2})}{(a_n+2)(a_{n-2}+2)}  - ①$ 
We can confirm the sign of $a_{n+2}-a_{n}$ and that of $a_{n}-a_{n-2}$ are same from ①. 
For $a_1 < a_3$ and $a_4<a_2$, we can say $a_{2m-1} < a_{2m+1}$ and $a_{2m+2}<a_{2m}$ respectively. 
Thus, $a_1<a_3<…<a_{2m-1}<…<2$, $a_2>a_4>…>a_{2m>}…>2$
$\{a_{2m-1}\}$ has upper bound and let $\alpha$ be its limit value. Also, $\{a_{2m}\}$ has lower bound and let $\beta$ be its limit value. 
We can gain $\beta=1+\frac{2}{\alpha}-②$ and $\alpha=1+\frac{2}{\beta}-③$ from $a_{2m}=1+\frac{2}{a_{2m-1}}, a_{2m+1}=1+\frac{2}{a_{2m}}(m\rightarrow\infty)$
Finally, we can get $\alpha=\beta=2$ from ② and ③. This completes the proof.
