Comparing $2$ infinite continued fractions 
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$

I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{A} \\
B = 2 + \dfrac{1}{B}$
Converting to quadratic equations:
$A^2 - A - 1 = 0 \\
B^2 -2B - 1 = 0$
Resulting to:
$2A = 1 + \sqrt{5} > 1 + \sqrt{2} = B$
My Question is:
Are there any more ways to solve this type of problem?

An alternative approach 
$$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} <  1 +\dfrac{1}{1} =2$$
$$B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}} > 2$$
showing $$A < B$$
Thus
$$2A = 2 +\dfrac{2}{A} > 2 +\dfrac{2}{2} =3$$
$$B = 2 +\dfrac{1}{B} <  2 + \dfrac12  $$
showing $$2A > B$$
