Hello Ernesto,
The minimal polynomial having $$1+\sqrt{3}$$ as a root is:
$$x^2-2x-2=0$$
Using this as our characteristic equation, we obtain the recursion:
(1) $$A_{n+1}=2A_{n}+2A_{n-1}$$
Where we will have:
$$\lim_{n\to\infty}\left(\frac{A_{n+1}}{A_{n}} \right)=\sqrt{3}+1$$
Therefore, we may approximate $\sqrt{3}$ with:
$$\sqrt{3}\approx\frac{A_{n+1}}{A_{n}}-1=\frac{A_{n+1}-A_{n}}{A_{n}}$$
Using our recursion in (1), we may write:
$$\sqrt{3}\approx\frac{A_{n}+2A_{n-1}}{A_{n}}=1+\frac{1}{\dfrac{A_{n}}{2A_{n-1}}}$$
Now, since the recursion in (1) may be written in $A_{n}$ as:
$$A_{n}=2A_{n-1}+2A_{n-2}$$
We may write:
$$\sqrt{3}\approx1+\frac{1}{\dfrac{2A_{n-1}+2A_{n-2}}{2A_{n-1}}}=1+\frac{1}{1+\dfrac{1}{\dfrac{A_{n-1}}{A_{n-2}}}}$$
Using the recursion for $$A_{n-1}$$ we may state:
$$\sqrt{3}\approx1+\frac{1}{1+\dfrac{1}{\dfrac{2A_{n-2}+2A_{n-3}}{A_{n-2}}}}=1+\frac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{A_{n-2}}{2A_{n-3}}}}}$$
Hence, we will find that by allowing $n\to\infty$ and repeating the above process ad infinitum, we will obtain:
$$\sqrt{3}=1+\dfrac{1}{1+\dfrac{1}{2+ \dfrac{1}{1+\dfrac{1}{2+\cdots}}}}$$
Written in linear form, we have:
$$\sqrt{3}=\left[1;\overline{1,2} \right]$$