<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] You can work out a continued fraction form for $2A$ and compare it with $B$.
Start with
$A=1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}...}}}$
Then we multiply by 2, by multiplying the initial $1$ by $2$ and dividing the denominator by $2$ at the first layer:
$2A=2+\dfrac{1}{(1/2)(1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}...}})}$
Now do the indicated multiplication by $1/2$ in the analogous way:
$2A=2+\dfrac{1}{(1/2)+\dfrac{1}{\color{blue}{(2)(1+\dfrac{1}{1+\dfrac{1}...}})}}$
The blue expression just gives us back our original $2A$, so this pattern repeats giving us:
$2A=\color{blue}{2}+\dfrac{1}{\color{brown}{(1/2)}+\dfrac{1}{\color{blue}{2}+\dfrac{1}{\color{brown}{(1/2)}+\dfrac{1}{...}}}}$
Versus:
$B=\color{blue}{2}+\dfrac{1}{\color{brown}{2}+\dfrac{1}{\color{blue}{2}+\dfrac{1}{\color{brown}{2}+\dfrac{1}{...}}}}$
Now note the coloring I gave to the summands.  By considering the convergent one can see that when the numerators are all $1$, the value of a continued fraction increases when the blue entries are increased but decreases when the brown entries are increased, provided these entries are all positive.  Here the blue entries for $2A$ and $B$ are the same but the brown entries are greater for $B$, therefore $2A>B$.
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