Do I need to analyse sequence given by $ x_{1+n} = \frac{1}{2 + x_{n}}$ without an equation with $0$? I have a problem with exercises with sequences given by recursion when I need to "prove convergence and find limit if it exists" and I am given a recursion of that kind:
$$ x_{1+n} = \frac{1}{2 + x_{n}}, x_1 \in (0 ; \infty)$$
It is fairly easy to find the limit - I just assume that the limit exists in $ \mathbb{R}$ and then use arithmetic properties of limits:
$$\lim_{n \to \infty} x_{n+1} = \lim_{n \to \infty} x_{n}$$
$$\lim_{n \to \infty} x_{n} = l, l \in \mathbb{R}>0$$
Taking my recursion:
$$l = \frac{1}{2 + l}$$
$$l^2 +2l - 1 = 0$$
$$l_1 = \sqrt{2} - 1 \in D$$
$$l_2 = -1 - \sqrt{2} \notin D$$
So my only possible limit in $ \mathbb{R}$ is $l = \sqrt{2} - 1$. That is if I can actualy prove that the limit exists - that is: the sequence is monotonous and bounded. And here is my problem - it is just impossible to analyse without computer the difference of:
$$x_{1+n} - x_{n} = \frac{1}{2 + x_{n}} - x_{n}$$
In search of limits I just multiply both sides of equation by $ \lim_{n \to \infty} x_{n} = l$ and it is impossible to do so here, so I get:
$$x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$$
Then I can't tell when it is bigger than $0$ to analyse monotonicity and I can't the see for which values o $n$ which values of $n+1$ i get (to get the boundary) because min value gets crazy.
So I just waneted to ask - am I missing something? Is it possible to make here $x_{1+n} - x_{n} = \frac{-x_{n}^2-2x_n+1}{2 + x_{n}}$ an equality with $0$ and analyse simpler function (red one on the picture)?

 A: $$X_{n+1}=\frac{1}{2+X_n} \implies 2 X_{n+1}+X_{n+1}X_n=1$$
Let $X_n=\frac{Y_{n-1}}{Y_n}$, then
$$2 \frac{Y_{n}}{Y_{n+1}}+\frac{Y_n}{Y_{n+1}}\frac{Y_{n-1}}{Y_n}=1 \implies 2Y_n+Y_{n-1}=Y_{n+1}.$$
Let $Y_n=t \implies t^2-2t-1=0 \implies t=1\pm \sqrt{2}.$
So $$Y_n=p(1+\sqrt{2})^n+q (1-\sqrt{2})^{n} $$
$$\implies X_n=\frac{(1+\sqrt{2})^{n-1}+r(1-\sqrt{2})^{n-1}}{(1+\sqrt{2})^{n}+r(1-\sqrt{2})^{n}}, r=q/p.$$
$$\lim_{n \to \infty}X_{\infty}=\frac{1}{1+\sqrt{2}}=\sqrt{2}-1$$
A: Although $x_1$ can be any positive number, all the terms starting from $x_2$ are less than $\frac 12$, so cannot be far from your limit.  An approach that can be useful is to write one term as the limit plus an error term, so here let $x_i=\sqrt 2-1+\epsilon$  Then
$$x_{i+1}=\frac 1{2+x_i}=\frac 1{1+\sqrt 2 + \epsilon}\\
x_{i+1}=\frac{\sqrt 2-1}{1+(\sqrt 2-1)\epsilon}\\
x_{i+1}\approx (\sqrt 2-1)-(\sqrt 2-1)^2\epsilon$$
where I have used the first order approximation to $\frac 1{1+\epsilon}$.  We see from this that the error is decreased by a factor about $6$ every step, so the sequence will converge.  To be more formal, you can bound the error from above using the fact that $x_i \in (0,\frac 12)$.  You won't get this fast a decrease, but any factor less than $1$ is good enough.
A: This is a Möbius transformation. Once you get the roots $l_1, l_2$ of the characteristic function $l^2+2l-1=0$, it follows that $1-2l_1=l_1^2$ and $1-2l_2=l_2^2$. Then
$$
x_{n+1}-l_1 = \frac{1}{2+x_n}-l_1 = \frac{1-2l_1-l_1 x_n}{2+x_n} = \frac{l_1^2-l_1 x_n}{2+x_n} = -l_1 \frac{x_n-l_1}{2+x_n} \tag 1
$$
Similarly
$$
x_{n+1}-l_2 = -l_2 \frac{x_n-l_2}{2+x_n} \tag 2
$$
$(1) \div (2)$ (you can do this because $x_n>0>l_2$),
$$
\frac{x_{n+1}-l_1}{x_{n+1}-l_2} = \frac{l_1}{l_2}\cdot \frac{x_n-l_1}{x_n-l_2}
$$
Therefore $\frac{x_n-l_1}{x_n-l_2}$ is a geometric sequence,
$$
\frac{x_n-l_1}{x_n-l_2} = \left(\frac{l_1}{l_2} \right)^{n-1} \cdot \frac{x_1-l_1}{x_1-l_2} \tag3
$$
Then $$x_n=\frac{l_1-\frac{x_1-l_1}{x_1-l_2}\left( \frac{l_1}{l_2}\right)^{n-1} \cdot l_2}{1- \frac{x_1-l_1}{x_1-l_2}\left(\frac{l_1}{l_2}\right)^{n-1}}$$
As $n\to \infty, \left(\frac{l_1}{l_2}\right)^{n-1} \to 0, x_n \to l_1 = \sqrt 2 - 1$.
To solve using matrices, see here for an example.
