Relation between two rational sequences approximating square root 2 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. 

Let $\left(\frac{1-\sqrt 2}{1+\sqrt 2}\right)^{2^n} = t$ for simplicity. Then, 
$$\frac{b_{n+1}-\sqrt 2}{b_{n+1}+\sqrt 2} = t$$ implies
\begin{align}
b_{n+1} &= \sqrt 2 \frac{1+t}{1-t} \\
&=\sqrt2 \frac{(1+\sqrt 2)^{2^n} +(1-\sqrt 2)^{2^n}}{(1+\sqrt 2)^{2^n} - (1-\sqrt 2)^{2^n}}\\
&= a_{2^n}
\end{align}
