Limit of a Sequence with odd terms increasing even terms increasing Let $x_0=0$ and $x_{n+1}=\frac{1}{2+x_n}$ for all $n\in \mathbb{N}$. Show the limit exists and find its value.
Here we have a sequence with odd terms increasing even terms increasing.
We also have $0 \leq x_n \leq 1$.
So I think the sequence converges in this case. But I'm not sure how to find the limit.
Any help will be appreciated!
 A: Here is a different proof to Igors.

Let us look at the following difference
$$\|x_{n+1} - x_n\| = \left\|\frac{1}{2+x_n} - \frac{1}{2+x_{n-1}}\right\| = \left\|\frac{x_{n-1}-x_n}{(2+x_n)(2+x_{n-1})}\right\| \le \frac{1}{4}\|x_{n-1}-x_n\|.$$
Where the last inequality follows from the fact that $x_n \ge 0$ for all $n$.
Thus, we see that the sequence is convergent.
For the limit $x^*$ we have that
$$ x^* = \frac{1}{2+x^*}.$$
Solving the above we see that $x^* = -1 \pm \sqrt{2}$. Thus, our limit point is $\sqrt{2}-1$.

The above argument at a high level uses the idea that the map $T(x) = \frac{1}{2+x}$ is a contraction map on $[0, \infty)$. Hence has a unique fixed point.
A: Step 1: Assume the limit exists, and is equal to $L.$ Then $L = \frac{1}{2+L}.$ I assume you can solve that equation. Call the solution $L.$ Now, if $x_n = L + \epsilon,$ show that $|x_{n+1} - L| < \epsilon.$ It is tedious, but works. A slicker way is to express $x_{n+2}$ in terms of $x_n,$ and note that this is given by a linear fractional transformation, which is given by a $2\times2$ matrix, so the proof is easy. However, I assume this is not something you are supposed to know, so do it in the tedious way.
A: Let $L=\sqrt{2}-1$. It is the only nonnegative root of the equation $x=\frac{1}{2+x}$.
Clearly the sequence $\{x_n\}$ is bounded between $0$ and $1$. If it does not converge to $L$, it will have a subsequence that is bounded away from $L$. In turn, some subsequence of this subsequence will converge to some $L'\in[0,1]$ (because $[0,1]$ is compact) but $L'\ne L$ (because this subsequence is bounded away from $L$). Consequently, $L'$ is a second nonnegative root of $x=\frac{1}{2+x}$. This is impossible. Hence $\{x_n\}$ must converge to $L$.
