We wanna show that the sequence $(t_n)_{n=1}^{\infty}$ recursively defined by $t_1=1$, $t_{n+1}=\frac{1}{2+t_n}$ for $\forall n \in \mathbb{N}$ converges to $-1+\sqrt 2$.
NOTE: i have to use the CONTRACTION THEOREM TO SHOW THIS. I have to make the recursively sequence a $f(x)$ and SHOW that it is a contraction.
Now i started on this problem and i used according to the Contraction Theorem that
For every $x\subset$ $I$ where $I=[0,1]$ we have that $f(x)\subset I$
And that there is a number $c$ such that $0\leq c<1$ such that for every $x,y$ in the interval we have $|f(x) - f(y)|\leq c|x-y|$
Now i started of and then after some algebra i got $\leq \frac{|x-y|}{|2+x| |2+y|}$
but now should i take $x=0, y=0$ because that is the minimim of the interval and i will get $c=\frac{1}{4}$? But am left with the fraction $\frac{1}{4} |x-y|$ which is also true according to the definition but i dont understand that why $|x-y|$ in the numerator still can stand after we decided that in the denominator we plugged in $x=0, y=0$
Use MVT instead to find $0\leq c<1$, i.e. $\exists \varepsilon \in (x,y) \subset [0,1]$ such that $$\left|f(y)-f(x)\right|=\left|f'(\varepsilon)\right|\cdot\left|y-x\right|$$ where $f(x)=\frac{1}{2+x}$. Or $$\left|f(y)-f(x)\right|=\left|\frac{1}{(2+\varepsilon)^2}\right|\cdot\left|y-x\right|=\frac{1}{(2+\varepsilon)^2}\cdot\left|y-x\right| \tag{1}$$ and since $0\leq \varepsilon \leq 1 \Rightarrow 2\leq 2+\varepsilon \leq 3 \Rightarrow 4\leq (2+\varepsilon)^2 \leq 9 \Rightarrow \frac{1}{4}\geq \frac{1}{(2+\varepsilon)^2} \geq \frac{1}{9}$. As a result, $(1)$ becomes $$\left|f(y)-f(x)\right|\leq\frac{1}{4}\cdot\left|y-x\right| \tag{2}$$ for $\forall x,y \in [0,1]$. So we take $c=\frac{1}{4}$.
Now, let's show that $$\forall x \in [0,1] \Rightarrow f(x) \in [0,1] \tag{3}$$
$0\leq x \leq1 \Rightarrow 2\leq 2+x \leq3 \Rightarrow \frac{1}{2}\geq \frac{1}{2+x} \geq \frac{1}{3} \Rightarrow 1 > \frac{1}{2} \geq f(x) \geq \frac{1}{3}>0$.
From $(2)$ and $(3)$ we have that $f(x)$ a contraction mapping on $[0,1]$. According to Banach fixed-point theorem and the fact that $t_1=1 \in [0,1]$, there $\exists! t^{*} \in [0,1]$ such that $t^{*}=f(t^{*})$ where $\lim\limits_{n\rightarrow\infty}t_n=t^{*}$ or $$t^{*}=\frac{1}{2+t^{*}} \Leftrightarrow (t^{*})^2+2t^{*}-1=0$$ This has 2 solutions $-1+\sqrt{2}$ and $-1-\sqrt{2}$, but only $-1+\sqrt{2} \in [0,1]$