TITLE: Relation between two rational sequences approximating square root 2
QUESTION [9 upvotes]: We define recursive sequences $a_{n+1}=1+\frac 1{1+a_n}$, $a_1=1$ and $b_{n+1}=\frac{b_n^2+2}{2b_n}$, $b_1=1$. I wish to show that $b_{n+1}=a_{2^n}$.
This can be proven using closed forms expressions related to continued fractions. I know that $a_n$ can be expressed as $$a_n=\sqrt2\cdot  \frac{(1+\sqrt 2)^n +(1-\sqrt 2)^n}{(1+\sqrt 2)^n - (1-\sqrt 2)^n}$$
On the other hand, we can prove inductively that $$\frac{b_{n+1}-\sqrt 2}{b_{n+1}+\sqrt 2}=\left(\frac{1-\sqrt 2}{1+\sqrt 2}\right)^{2^n}$$
So the relation $a_{2^n}=b_{n+1}$ can be deduced by expanding the fractions. However the computation is rather tedious, I am looking for a proof that does not involve expanding everything into closed form expressions. Thanks.

REPLY [4 votes]: Here is an outline of a proof by induction that doesn't use the closed form expressions.
We are given $a_{n+1}=\dfrac{2+a_n}{1+a_n}$ and $a_1=1$.  
First prove by induction on $m$ that $a_{n+m}=\dfrac{2+a_ma_n}{a_m+a_n}.$
It follows that $a_{2n}=\dfrac{2+a_n^2}{2a_n}$.  
Now prove by induction that $b_{n+1}=a_{2^n}$.