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?


Are there any more ways to solve this type of problem?

We can get some loose bounds by truncating the fractions.
$A = 1 +\dfrac{1}{1 + \left[\frac{1}{1 + \frac{1}{\ddots}}\right]}$
The quantity in the square brackets is clearly smaller than 11 (we have 11 divided by something larger than 11); it follows that $A>1+\frac{1}{1+1}=\frac{3}{2}$ 
Similarly, we have
$B = 2 +\dfrac{1}{2 + \left[\frac{1}{2 + \frac{1}{\ddots}}\right]}$
Here, we use the fact that the bracketed quantity is greater than zero to get $B<2+\frac{1}{2} = \frac{5}{2}$
Hence $\boxed{2A>B}$
