Continued fraction of $ \sqrt{3} $ How can I solve the continued fraction of $\sqrt{3}$?
I know that a continued fraction is [1;1,2,1,2,1,2,....] -> this is the continued fraction of $\sqrt{3}$, but how can I solve? What are the steps? 
I don't understand this proof.

Thank you so much.

We know $\sqrt{3}$ lies between 1 and 2 (because $1^2 < 3 < 2^2$). So the first digit of the continued fraction expansion is 1. So
$\sqrt{3} = 1 + (\sqrt{3}-1)$
and $\sqrt{3}-1$ must be between $0$ and $1$. So next we want to find the reciprocal of $\sqrt{3}-1$:
$\frac{1}{\sqrt{3}-1} = \frac{\sqrt{3}+1}{(\sqrt{3}-1)(\sqrt{3}+1)} = \frac{\sqrt{3}+1}{2}$
$\frac{\sqrt{3}+1}{2}$ lies between $1$ and $2$ so the next term in the continued fraction is 1. So we have $\sqrt{3} = [1;1,\dots]$.
The remaining amount is $\frac{\sqrt{3}+1}{2}-1 = \frac{\sqrt{3}-1}{2}$ so next we want to find the reciprocal of $\frac{\sqrt{3}-1}{2}$:
$\frac{2}{\sqrt{3}-1} = \frac{2(\sqrt{3}+1)}{2} = \sqrt{3}+1$
$\sqrt{3}+1$ is between 2 and 3, so the next term in the continued fraction is 2. So we have $\sqrt{3} = [1;1,2,\dots]$. 
The remaining amount is $(\sqrt{3}+1)-2=\sqrt{3}-1$. We could continue, but we have already calculated the reciprocal of $\sqrt{3}-1$ above, so now the continued fraction just repeats and we have:
$\sqrt{3} = [1;1,2,1,2,1,2, \dots]$
An alternative way of seeing why this continued fraction repeats is to note that
$(\sqrt{3}+1)^2 = 2\sqrt{3} + 4 = 2(\sqrt{3}+1) + 2$
so $\sqrt{3}+1$ is a solution to $x^2=2x+2$, which we can re-write as:
$x = 2 + \dfrac{2}{x} = 2 + \dfrac{1}{\dfrac{x}{2}} = 2 + \dfrac{1}{1 + \dfrac{1}{x}}$
so $\sqrt{3} + 1 = [2;1,2,1,2,1,\dots]$.
