Showing the sequence converges $a_{1}=\frac{1}{2}$, $a_{n+1}=\frac{1}{2+a_{n}}$ Showing the sequence converges $a_{1}=\frac{1}{2}$,  $a_{n+1}=\frac{1}{2+a_{n}}$.
I already know that if $(a_{n})$ converges then it does to  $\sqrt{2}-1$.But i dont't know how to prove that this sequence cenverges.
EDIT
I think that the subsequence $(a_{2n+1})$ is monotonic decreasing and the subsequence $(a_{2n})$ is monotonic increasing.
 A: Indeed, this sequence is not monotonic. 
It seems that for $$n = 2k, k \in \mathbb{N}, a_n \geq \sqrt{2} - 1$$ and for $$n = 2k+1, k \in \mathbb{N}, a_n \leq \sqrt{2}-1$$ (this can be proved using induction).
Now, $$a_{2n+2} - a_{2n} = \displaystyle \frac{1}{2+a_{2n+1}} - a_{2n} = \frac{1}{2+\frac{1}{2+a_{2n}}}  - a_{2n} = \frac{-2a_{2n}^2-4a_{2n}+2}{5+2a_{2n}}.$$ 
Because $a_{2n} \geq \sqrt{2}-1 \implies -2a_{2n}^2-4a_{2n}+2 \leq 0$ so $(a_{2n})_{n \in \mathbb{N}}$ is decreasing.
You can prove analogously that $(a_{2n+1})_{n \in \mathbb{N}}$ is increasing.
Now you write the recursion relation for $a_{2n}$ and $a_{2n+1}$, apply $\displaystyle \lim_{n \to \infty}$, and you get that the limit is $\sqrt{2}-1$.
You can also just apply $\displaystyle \lim_{n \to \infty}$ in the starting recursion relation, because $(a_{2n})_{n \in \mathbb{N}}$ decreases to $\sqrt{2} - 1$ and $(a_{2n+1})_{n \in \mathbb{N}}$ increases towards $\sqrt{2}-1$, but writing it for $n$ even and $n$ odd is more rigurous.
$$ a_{2n+2} = \displaystyle \frac{1}{a_{2n+1}+2} = \frac{1}{2+\frac{1}{2+a_{2n}}} .$$ Applying $\displaystyle \lim_{n \to \infty}$, we get that $l = \displaystyle \frac{1}{2+\frac{1}{2+l}} \iff 2l^2 + 4l - 2 = 0$, and the equation has the solutions $l_1 = -1-\sqrt{2}$ and $l_2 = -1+\sqrt{2}$, but since $a_n >0,\forall n \in \mathbb{N}$, the limit is $l = -1+\sqrt{2}$. This is the same for odd $n$.
A: The approach from https://math.stackexchange.com/a/124288/42969 works
here as well:
We have
$$
 | a_{n+2} - a_{n+1}| = \frac{|a_{n+1}-a_n|}{(2+a_{n+1})(2+a_n)}
 \le \frac 14 |a_{n+1}-a_n|
$$
which implies that $(a_n)$ is a Cauchy sequence and therefore convergent.

And using the approach from https://math.stackexchange.com/a/133878/42969
one can obtain an explicit formula for $a_n$:
Let $u = \sqrt 2 -1 > 0$ and $v = -\sqrt 2 - 1 < 0$ be the solutions
of $x^2 + 2x - 1 = 0$. Then an elementary calculation gives
$$
 \frac{a_{n+1} - u}{a_{n+1}-v} = \frac uv \frac{a_n - u}{a_n-v}
$$ 
and therefore
$$
  \frac{a_{n} - u}{a_{n}-v} = \left(\frac uv\right)^{n-1} \frac{a_1 - u}{a_1-v}
$$
The quotient $u/v$ is negative and its absolute value is less than one.
It follows that $a_n \to u$  (and also that the differences $a_n - u$
are alternating positive and negative).
A: Another look at the problem from the contraction mapping point of view. We will look at the function $f(x)=\frac{1}{2+x}$ s.t. $a_{n+1}=f(a_n)$.


*

*$\forall x \in \left[0,\frac{1}{2}\right] => f(x) \in \left[0,\frac{1}{2}\right]$. Indeed 
$$0\leq x\leq \frac{1}{2} \Rightarrow 2\leq 2+x\leq 2+\frac{1}{2} \Rightarrow \frac{1}{2}\geq \frac{1}{2+x}\geq \frac{1}{2+\frac{1}{2}}>0$$ or $$\frac{1}{2}\geq f(x)\geq 0$$

*$f(x)$ is a contraction mapping on $\left[0,\frac{1}{2}\right]$ (it's wider, but we don't need more than this), from MVT $\forall x,y \in \left[0,\frac{1}{2}\right], \exists c$ in between them s.t.
$$|f(x)-f(y)|=|f'(c)||x-y|=\left|-\frac{1}{(2+c)^2}\right||x-y|\leq\frac{1}{4}|x-y|$$


Since $a_1 \in \left[0,\frac{1}{2}\right]$, from Banach fixed-point theorem, the sequence has a limit on $\left[0,\frac{1}{2}\right]$ which you can find from $L=\frac{1}{2+L}$ (which, I believe, is what you did).
A: Since you already have a reasonable conjecture consider the sequence $(b_n)_{n\geq1}$ defined by
$$a_n=\sqrt{2}-1+b_n,\qquad{\rm resp.,}\qquad b_n=a_n+1-\sqrt{2}\qquad(n\geq1)\ ,$$
and try to prove that $\lim_{n\to\infty} b_n=0$, which should be simpler.
A: Let $b_n:=a_{2n-1}$ and $c_n:=a_{2n}$ for $n\in\mathbb{N}$. Since $a_n$ is bounded because $a_n\leqslant 1/2$ and $a_n>0$ for all $n$, so are the sequences $b_n$  and $c_n$. Now we show that $b_n$ is monotonic decreasing while $c_n$ is monotonic increasing. 
First note that $b_n>\sqrt{2}-1$ for all $n$. For $n=1$ it is clear since $b_1=a_1=1/2>\sqrt{2}-1$. Then 
$$b_{n+1}=a_{2n+1}=\frac{1}{2+a_{2n}}=\frac{1}{2+\frac{1}{2+a_{2n-1}}}=\frac{2+a_{2n-1}}{5+2a_{2n-1}}=\frac{2+b_{n-1}}{5+2b_{n-1}}\\=\frac{2+b_{n-1}}{1+2(2+b_{n-1})}>\frac{2+(\sqrt{2}-1)}{1+2(2+\sqrt{2}-1)}\\=\frac{1+\sqrt{2}}{1+2(1+\sqrt{2})}=\frac{1}{(\sqrt{2}-1)+2}=\frac{1}{\sqrt{2}+1}=\sqrt{2}-1$$
where we have used the fact that the function $$f(x):=\frac{x}{1+2x}$$ is increasing on $\mathbb{R}^+$ together with the inductive hypothesis $b_{n-1}>\sqrt{2}-1$. Next we show that $b_n$ is monotone decreasing 
Note
$$b_{n+1}-b_n=a_{2n+1}-a_{2n-1}=\frac{1}{2+a_{2n}}-a_{2n-1}=\frac{1}{2+\frac{1}{2+a_{2n-1}}}-a_{2n-1}\\=\frac{2+a_{2n-1}}{5+2a_{2n-1}}-a_{2n-1}=\frac{2(1-2a_{2n-1}-a_{2n-1}^2)}{5+2a_{2n-1}}\\=\frac{2(1-2b_{n-1}-b_{n-1}^2)}{5+2b_{n-1}}=\frac{2(\sqrt{2}-1-b_{n-1})(\sqrt{2}+1+b_{n-1})}{5+2b_{n-1}}<0$$
since by previous observation $b_n>\sqrt{2}-1$ for all $n$. Therefore $b_n$ is bounded and monotone so by Bolzano-Weierstrass we obtain that $b_n$ has a limit point $b$. Hence
$$\lim_nb_{n+1}=\lim_n\frac{2+b_{n-1}}{5+2b_{n-1}}\Rightarrow b=\frac{2+b}{5+2b}\Rightarrow b^2+2b-1=0\Rightarrow b\in\{\sqrt{2}-1,-\sqrt{2}-1\}$$
Since $b_n>0$ for all $n$ then $b=\sqrt{2}-1$. 
Exaclty analogue arguments for the sequence $c_n$ one finds that it is bounded and monotone increasing satisfying $c_n<\sqrt{2}-1$ for all $n$. The limit point of $c_n$ we denote by $c$ can be shown to be $\sqrt{2}-1$. 
Finally note by construction of $b_n$ and $c_n$ it must be the case that 
$$\lim_n b_n=\lim\sup_n a_n\hspace{0.2cm}\text{and}\hspace{0.2cm} \lim_n c_n=\lim\inf_n a_n$$
Since $\lim_n b_n=\sqrt{2}-1=\lim_n c_n$ we get
$$\lim_n a_n=\lim\sup_n a_n=\lim\inf_n a_n=\sqrt{2}-1$$
A: Here is another proof.
Say $$a_n = \frac{p_n}{q_n}$$
then from the identity we have that
$$p_{n+1} = q_n, q_{n+1} = 2q_n + p_n = 2q_n + q_{n-1}$$
$q_n$ are of the form
$$ A (\sqrt{2} + 1)^n + B (1 - \sqrt{2})^n $$
($\sqrt{2} + 1$ and $1 - \sqrt{2}$ are roots of $x^2 = 2x + 1$)
Thus $$a_n = \frac{q_{n-1}}{q_n}  = \frac{A (\sqrt{2} + 1)^{n-1} + B (1 - \sqrt{2})^{n-1}}{A (\sqrt{2} + 1)^n + B (1 - \sqrt{2})^n}\to \frac{1}{1 + \sqrt{2}} = \sqrt{2} -1 $$
(Divide both numerator and denominator by $(\sqrt{2} + 1)^n$)
