Proof that if $a_1=1$ and $a_{n+1}=1+\frac{1}{1+a_n}$ Question: Prove that if 
$$a_n=\left\{
  \begin{array}{ll}
    a_1=1\\
    a_{n+1}=1+\frac{1}{1+a_n}
  \end{array}
\right.$$
then $a_n$ converges, and then find $\lim_{n \to \infty}a_n$.
I found that $\lim_{n \to \infty}a_n=\sqrt{2}$, and that $1\leq a_n<2$, but the convergence seems difficult.
My idea is to show that $a_{2n}$ is decreasing, and $a_{2n-1}$ is increasing,  but I don’t see how.
I’d be thankful for any hint.

Most inelegant method I guess:
$a_{n+2}=1+\cfrac{1}{1+a_{n+1}}=1+\cfrac{1}{1+1+\cfrac{1}{1+a_n}}=1+\cfrac{1+a_n}{3+2a_n}
$
So $a_{n+2}-a_n=2\cfrac{2-a_n^2}{3+2a_n}$
Clearly from definition $a_n>1$
Also if $a_n>\sqrt{2}$ then $a_{n+2}>1+\cfrac{1+\sqrt{2}}{3+2}=\cfrac{6}{5}+\cfrac{\sqrt{2}}{5}>\cfrac{4\sqrt{2}}{5}+\cfrac{\sqrt{2}}{5}=\sqrt{2}$
if $a_n<\sqrt{2}$ then $a_{n+2}<1+\cfrac{1+1}{3+2\sqrt{2}}=1+\cfrac{2(3-2\sqrt{2})}{9-8}=7-4\sqrt{2}=5\times\cfrac{7}{5}-4\sqrt{2}<5\times \sqrt{2}-4\sqrt{2}=\sqrt{2}$
Now $a_1=1<\sqrt{2}$, $a_2=1.5>\sqrt{2}$ So by induction, the sequences $a_1,a_3,a_5,\ldots$ and $a_2,a_4,\ldots$ are monotone increasing/decreasing respectively, and is bounded above/below by $\sqrt{2}$. These two sequences clearly converges to the same limit, and so $(a_n)$ converges to the common limit which is $\sqrt{2}$  (so in fact $\sqrt{2}$ is supremum and infimum of the sequences)
