$x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}$ is convergent if $x_0 > 0$ and $x_1 > 0$ I have stumbled across this problem in an olympiad-material sent by my teacher. Prove that the sequence $$x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}$$ is convergent when $x_0 > 0$ and $x_1 > 0$. My only idea which has absolutely failed so far is trying to prove that the sequence is monotone and bounded, but after some examples, I found that it isn't monotone. Could you help and maybe suggest other strategies for proving the convergence of a sequence other than the one I tried?
 A: 
$x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}$ is convergent if $x_0 > 0$ and $x_1 > 0$.

Proof: an initial guess for the limit of $x_n$, is $1$ by replacing $x_n$, $x_{n+1}$ and $x_{n+2}$ with $l$ in the recurrence relation and solving for $l$. We hence prove that $x_n\to 1$. Note that if $\epsilon<x_n<\frac{1}{\epsilon}$ and $\epsilon<x_{n+1}<\frac{1}{\epsilon}$ for some $0<\epsilon<1$, then
$$
\epsilon<
\frac{2+\epsilon}{2+\frac{1}{\epsilon}}<
x_{n+2}=\frac{2+x_{n+1}}{2+x_n}
<\frac{2+\frac{1}{\epsilon}}{2+\epsilon}<\frac{1}{\epsilon},
$$
where the inequalities $\frac{2+\frac{1}{\epsilon}}{2+\epsilon}<\frac{1}{\epsilon}$ and $\epsilon<
\frac{2+\epsilon}{2+\frac{1}{\epsilon}}$ follow from $0<\epsilon<1$ immediately. Hence, by defining $\epsilon=\min\{x_0,x_1,\frac{1}{2}\}$, we have proven that there exists $0<\epsilon<1$ such that $x_n>\epsilon$ for all $n$.
The next steps follow as below:
$${
x_{n+2} = \frac{2+x_{n+1}}{2+x_{n}}\implies
\\
x_{n+2}-1 = \frac{x_{n+1}-x_n}{2+x_{n}}\implies
\\
|x_{n+2}-1| = \frac{|x_{n+1}-x_n|}{2+x_{n}}=\frac{|x_{n+1}-1-(x_n-1)|}{2+x_{n}}\implies
\\
|x_{n+2}-1| \le \frac{|x_{n+1}-1|+|x_n-1|}{2+x_{n}}\implies
\\
|x_{n+2}-1| \le \frac{|x_{n+1}-1|+|x_n-1|}{2+\epsilon}\implies
\\
|x_{n+2}-1|+\frac{1}{2}|x_{n+1}-1| \le \frac{(2+\frac{\epsilon}{2})|x_{n+1}-1|+|x_n-1|}{2+\epsilon}\implies
\\
|x_{n+2}-1|+\frac{1}{2}|x_{n+1}-1| \le \frac{|x_{n+1}-1|+\frac{1}{2+\frac{\epsilon}{2}}|x_n-1|}{\frac{2+\epsilon}{2+\frac{\epsilon}{2}}}\implies
\\
|x_{n+2}-1|+\frac{1}{2}|x_{n+1}-1| \le \frac{|x_{n+1}-1|+\frac{1}{2}|x_n-1|}{\frac{2+\epsilon}{2+\frac{\epsilon}{2}}}\implies
\\
}$$
By defining $y_n=|x_{n+1}-1|+\frac{1}{2}|x_n-1|$ and $r=\frac{2+\frac{\epsilon}{2}}{2+\epsilon}$, we have proven that $|y_{n+1}|\le r|y_n|$ and $0<r<1$. Hence $y_n\to 0$ which means that $x_n$ is convergent to $1$ and the proof is complete $\blacksquare$
