Show that this sequence converges. (cauchy criterion) Given $a_0 \geq 0$ and a sequence ($a_n)_{n\in\mathbb{N}}$

$$ a_{n+1}= \frac1{(2+a_{n})}.$$

for ${n\in\mathbb{N_0}}$.
Show that $(a_n)_{n\in\mathbb{N}}$ is convergent and determine the limit.
All I've got so far is that this sequence is not monotonous, but that does not help alot, does it?
 A: If we show that the sequence is convergent, then its limit is $L=\sqrt2-1$, since it should satisfy 
$$
L=\frac1{2+L}.
$$
It is clear that $a_n>0$ for all $n$. 
And we have
$$
|a_{n+1}-a_n|=\left|\frac1{2+a_n}-\frac1{2+a_{n-1}}\right\|=\frac{|a_{n-1}-a_n|}{(2+a_n)(2+a_{n-1})}\leq\frac{|a_{n}-a_{n-1}|}4.
$$
Inductively,
$$
|a_{n+1}-a_n|\leq\frac1{4^n}\,|a_1-a_0|.
$$
Then
$$
|a_{n+k}-a_n|\leq\sum_{j=1}^{k-1}|a_{n+j+1}-a_{n+j}|\leq|a_1-a_0|\,\sum_{j=0}^{k-1}4^{-j-n}=4^{-n}\,\frac{4(1-4^{-k})|a_1-a_0|}{3}.
$$
This shows that the sequence is Cauchy, and thus convergent. 
A: It is likely that an argument that uses the fact that $a_0,a_2,\dots$ and $a_1,a_3,\dots$ are monotone is intended. We give a more calculus-style argument. 
If the limit exists, let it be $b$. Then $b=\frac{1}{2+b}$, and therefore $b=\sqrt{2}-1$.
All the $a_n$ are $\ge 0$. 
Let $f(x)=\frac{1}{2+x}$. Then $f'(x)=-\frac{1}{(2+x)^2}$. Note that $f(b)=b$. 
Thus $a_{n+1}-b=f(a_n)-f(b)$
By the Mean Value Theorem 
$$f(a_n)-f(b)=(a_n-b)f'(c)$$
for some $c$ between $a_n$ and $b$. The derivative has absolute value $\lt \frac{1}{4}$. It follows that
$$|a_{n+1}-b|\le \frac{1}{4}|a_n-b|,$$
and therefore the sequence converges to $b$. 
Remark: If we wish to use the Cauchy Criterion, the same MVT argument will show that
$$|a_{n+2}-a_{n+1}|\le \frac{1}{4}|a_{n+1}-a_n|.$$ 
