Recursively defined sequence $a_{n+1} = \frac{1}{2 + a_{n}}$ Is the following recursively defined sequence
$$x_{n+1} = \frac{1}{2 + x_{n}}$$ for $ n \geq 0$  and  $x_{0} = 0$ convergent although it is not a monotonic sequence because $x_{1}=0.5, x_{2}=0.4$ and $x_{3} \approx 0.416$.
 A: The sequence is bounded from bellow(you can prove by induction the it is non negative)
The function $f(x)=\frac{1}{2+x}$ is a contraction map on $[0,+\infty]$,i.e it $|f(x)-f(y)| \leq M|x-y|$ where $0<M<1.$

Indeed $|f'(x)|=\frac{1}{(x+2)^2} \leq \frac{1}{2}$,so by Mean Value theorem is a contraction.

So the sequence $x_{n+1}=f(x_n)$ is a Cauchy sequence and has  limit(which is also the unique fixed point of $f$)because of the completeness of $[0,+\infty)$
A: If we apply the recurrence formula twice we get the recurrence
$$x_{n+1}=\frac1{2+\frac1{2+x_{n-1}}}=\frac{2+x_{n-1}}{5+2x_{n-1}}=\frac12-\frac1{10+4x_{n-1}}$$
with $x_0=0$ and $x_1=\frac12$. Then we can note that
$$x_{n-1}\in(-\sqrt{2}-1,\sqrt{2}-1)\implies x_{n+1}=\frac12-\frac1{10+4x_{n-1}}\in(-\sqrt{2}-1,\sqrt{2}-1)$$
$$x_{n-1}\in(\sqrt{2}-1,\infty)\implies x_{n+1}=\frac12-\frac1{10+4x_{n-1}}\in(\sqrt{2}-1,\infty)$$
So we can deduce that $x_{2k}\in(-\sqrt{2}-1,\sqrt{2}-1)$ for all $k\ge0$. We can then prove that the subsequence $x_{2k}$ is strictly increasing as
$$x_{n-1}\in(-\sqrt{2}-1,\sqrt{2}-1)\implies x_{n+1}=\frac12-\frac1{10+4x_{n-1}}\gt x_{n-1}$$
and similarly the subsequence $x_{2k+1}$ is strictly decreasing as $x_{2k+1}\in(\sqrt{2}-1,\infty)$ for all $k\ge0$ and
$$x_{n-1}\in(\sqrt{2}-1,\infty)\implies x_{n+1}=\frac12-\frac1{10+4x_{n-1}}\lt x_{n-1}$$
So by the Monotone convergence theorem the subsequence $x_{2k}$ converges and so does the subsequence $x_{2k+1}$ and they must in fact both converge to the fixed point $\sqrt{2}-1$ because of the nature of both sequences. Then we can just say that 
$$\left((x_{2k})_{k\in\mathbb{N}_0}\to\sqrt{2}-1\right)\land\left((x_{2k+1})_{k\in\mathbb{N}_0}\to\sqrt{2}-1\right)\implies(x_n)_{n\in\mathbb{N}_0}\to\sqrt{2}-1$$
A: By writing out a few terms we can see that the sequence changes monotony, or with derivative test we can see that it is breaking into sub sequences, 2k+1 and 2k that have different monotony. Seeing how they behave should lead to an equation for both sequences, which should give the same solution if the sequence converges.
