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?

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. 
