The limit of a simple sequence recursively defined 
Given a real number $a>0$, $(x_n)$ is recursively defined by 
  $$
\begin{cases}
x_1=a\\
x_{n+1}=\frac{2x_n+1}{x_n+2}
\end{cases}
$$
  Find a proof that $(x_n)$ is convergent and then find the limit.

Can anyone give me a hint on how to determine the limit of a sequence like that? I tried to write the first terms to see if I could find an equation that was not recursive for the $n$-th term, but I failed miserably. I dont really know how to find the answer to that without such equation.
Grateful for any help!
 A: HINT: if the sequence converges in $\Bbb R$ then applying limits both sides of the definition of $x_{n+1}$ we get
$$\lim_{n\to\infty} x_{n+1}=\lim_{n\to\infty} \frac{2 x_n+1}{x_n+2}\implies L=\frac{2L+1}{L+2}\implies L^2=1\implies L=1$$
where we had discarded $L=-1$ because the sequence $(x_n)$ is positive so it cannot converges to $-1$.
Then suppose that the limit of the sequence is $1$ and divide by cases. If $0<a<1$ then $x_1<x_2<1$, so you can try to setup a proof by induction to show that $x_n<1$ for all $n\in\Bbb N$ and that the sequence $(x_n)$ is strictly increasing.
If $a>1$ then $x_1>x_2$ and you can try to setup a proof by induction to show that the sequence $(x_n)$ is strictly decreasing and positive.
If the above fails for some case then you can try to show that the sequence doesnt converges to $1$, and because $1$ is the unique possible candidate for the limit then the sequence doesnt converge for the studied case.

An alternative method is studying the fixed points of $f(x):=\frac{2x+1}{x+2}$ in $(0,\infty)$. In this case we can see that $f$ is a contraction in this region (because $|f'(x)|<1$ for $x\in(0,\infty)$), so it converges to a unique fixed point in this region.
A: Coincidence! I just added a similar answer to another question using matrices.

Note: This is to answer your question about a non-recursive formula
  for the $n^{th}$ term (and consequently to find the limit). Also note that this is a very general approach for solving such problems.

$$x_n = \frac{p_n}{q_n}$$
where
$$ p_{n+1} = 2p_n + q_n$$
$$ q_{n+1} = p_n + 2q_n$$
[You can consider $s_n = p_n + q_n$ and do something clever, but we won't go there as we want a general approach]
Thus
$$A^n \begin{bmatrix}a\\1\end{bmatrix} = \begin{bmatrix}p_n\\q_n\end{bmatrix}$$
Where 
$$A = \begin{bmatrix}2&1\\1&2\end{bmatrix}$$
This matrix is diagonalizable and gives the general form of $p_n$ and $q_n$ to be
$$ u3^n + v (-1)^n$$
($3$ and $-1$ are the eigenvalues of $A$)
I will leave it to you to compute the the formulae for $p_n$ and $q_n$, which will prove convergence of $x_n$.
btw, the limit is $1$, because the matrix $A$ is symmetric. $A^n$ will be symmetric with entries of the form $x3^n + y(-1)^n$, and so the $u$ for $p_n$ will be the same as the $u$ for $q_n$ and the limit is their ratio.
A: Since $a_1>0$ the the recursive def'n of $a_{n+1}$ implies,by induction on $n,$ that $a_n>0$ for all $n.$ 
Let $a_n=1+d_n.$ We have $a_n>0$ so $ d_n>-1.$ By direct calculation we have $$d_{n+1}=a_{n+1}-1=[(2a_n+1)/(a_n+2)]-1=d_n/(3+d_n).$$
Now $d_n>-1\implies 3+d_n>2,$ so $$|d_{n+1}|=|d_n|/|3+d_n|\leq |d_n|/2.$$
The result $|d_{n+1}|\leq |d_n|/2$ implies that $d_n\to 0$ and hence $a_n\to 1.$
Note: The first sentence of the Answer by Masacroso shows you how to deduce that if $L=\lim_{n\to \infty}$ exists, then $L^2=1.$ And since $a_n>0$ for all $n,$ if $L$ exists then $L=1.$
A: When you have a sequence $x_{n+1}=f(x_n)$ with f(x) an increasing function ($f'(x)>0$).
You can prove the fact that the sequence has a ceiling (or a floor) and is increasing (or decreasing) by induction :
using the property of increasing function f : $a < b \Rightarrow f(a) < f(b)$
example :
if $x_{n} < L \Rightarrow f(x_{n})=x_{n+1} < f(L)=L$
the sequence has a ceiling. Provided that $x_1<L$
if $x_{n} < x_{n+1} \Rightarrow f(x_{n})=x_{n+1} < f(x_{n+1})=x_{n+2}$
the sequence is going up. Provided that $x_1<x_2$
Then you use the theorem : an increasing sequence with a ceiling necessarily converges. (or the other way)
