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?
On
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 $$|a_{n+2}-a_{n+1}|\le \frac{1}{4}|a_{n+1}-a_n|.$$
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.