<s>[INST] Prove that if $x_{n+2}=\frac{2+x_{n+1}}{2+x_n},$ then $x_n$ converges 
Let $x_0 > 0$, $x_1 > 0$ and $$x_{n+2} = \dfrac{2+x_{n+1}}{2+x_n}$$ for $n \in \{0,1,\dots\}$. Prove that $x_n$ converges.

My attempt:
If $x_n$ converges and $\lim\limits_{n\rightarrow\infty}x_n=a$ so $a=\frac{2+a}{2+a}$, which gives $a=1$. Now, 
$$x_{n+2}-1=\frac{x_{n+1}-x_n}{2+x_n}$$ 
and what is the rest? Thank you!
 [/INST] Clearly every $x_n$ is positive. With this we know that $$x_{n+2} = \frac{2 + x_{n+1}}{2 + x_n} = \frac{2}{2+x_n} + \frac{x_{n+1}}{2+x_n} < 1 + \frac{x_{n+1}}{2}.$$
Applying this recursively we get $$x_n < 1 + \frac{x_{n-1}}{2} < 1 + \frac{1 + \frac{x_{n-2}}{2}}{2} = 1 + \frac{1}{2} + \frac{x_{n-2}}{4} < $$
$$< 1 + \frac{1}{2} + \frac{1 + \frac{x_{n-3}}{2}}{4} = 1 + \frac{1}{2} + \frac{1}{4} + \frac{x_{n-3}}{8} < \frac{x_{n-k}}{2^k} + \sum_{i=0}^{k-1} \frac{1}{2^i}.$$
This bound is valid for any $k = 1 \ldots n$. For $k = n$ we get $$x_n < \frac{x_0}{2^n} + \sum_{i=0}^{n-1} \frac{1}{2^i} < \frac{x_0}{2^n} + 2. $$
Edit: This proves the sequence is bounded. But to get convergence we still need to prove something else, like it is monotone.  
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