<s>[INST] 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?
 [/INST] 
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 $\boxed{2A>B}$
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