proving convergence of $a_{n+1}=1+\frac{1}{1+a_{n}}$ 
$a_1=1.$ $a_{n+1}=1+\frac{1}{1+a_{n}}$

Prove that the sequence is convergent.
I'm trying to prove the convergence of this sequence but having trouble. At first I thought this might be a monotone sequence since then I can try monotone convergence theorem to prove its convergence.
But after checking some terms, I realized it seemed the sequence is oscillating. So I'm not sure how to prove the convergence of this sequence.
Thanks.
 A: This sequence is a Cauchy sequence so it converges.
First you see $a_n>0, \forall n \in \mathbb{N}$ from recursive relation.  [$a_1=1$ and $a_{n+1}$ is defined to added positive terms]
Second since $a_n>0$ thus $a_{n+1} = 1 + \frac{1}{1+a_n} \leq 2 $
Now consider
\begin{align}
|a_{n+1} - a_n| = \left| \frac{1}{1+a_n} - \frac{1}{1+a_{n-1}} \right|
 = \frac{|a_n - a_{n-1}|}{(1+a_n)(1+a_{n-1})} \leq \frac{1}{4} | a_n - a_{n -1}|
\end{align}
and this is cauchy sequence. [Series with this form is called contractive and after repeatively apply the same procedure continuing to $|a_2-a_1|$, and by Squeeze theorem you can easily guess $a_n$ is a Cauchy sequence]
In $\mathbb{R}$ cauchy sequence implies convergences so it converges. Then by taking limits $\lim_{n\rightarrow \infty} a_n = \alpha$ we have $\alpha^2 = 2$ and from $a_n>0$, $\alpha = \sqrt{2}$.
A: A method often useful with oscillating sequences: Let $b_n=|(a_n)^2-2|.$ Then $$0\le b_{n+1}=\frac {b_n}{(1+a_n)^2}\le \frac {b_n}{4} $$ because $1+a_n\ge 2$ by induction on $n$.
So $b_n$ decreases to  $0$. So $(a_n)^2\to 2$ with each $a_n>0.$
The motivation for the "$2$" in the definition of $b_n$ is that IF $a_n$ converges to a limit $L$ then $L=\lim_{n\to \infty}a_{n+1}=\lim_{n\to \infty} 1+\frac {1}{1+a_n}=1+\frac {1}{1+L},$ implying $L^2=2.$
