Comparing $2$ infinite continued fractions

Question

$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\ B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$

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?

Answer 1

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 $$2A > B$$

Answer 2

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 $\boxed{2A>B}$

Answer 3

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 $2A>B$.

Answer 4

Isn’t $\frac12B$ equal to $$ 1+\frac1{4+\frac1{1+\cdots}}\qquad <\qquad A $$ ?

Answer 5

Clearly $A,B,2B-A>0$.

Note that $B=2+\dfrac{1}{B}$ and $A=1+\dfrac{1}{A}\iff 2A=2+\dfrac{2}{A}$.

It follows, $2A-B=\dfrac{2}{A}-\dfrac{1}{B}=\dfrac{2B-A}{AB}>0$.