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.

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
$$\lim_n a_n=\lim\sup_n a_n=\lim\inf_n a_n=\sqrt{2}-1$$
