Proving continued fraction of $\sqrt{k^2 -1}$ is $<k-1, \overline{1, 2(k-1)}>$

Question

I know that $a_0$ must be $k-1$ since the integer part is $\lfloor \sqrt{k^2 -1} \rfloor$ but am struggling on the repeated part.

Answer 1

You now consider $$x_1=\frac1{\sqrt{k^2-1}-(k-1)}=\frac{\sqrt{k^2-1}+(k-1)}{k^2-1-(k-1)^2} =\frac{\sqrt{k^2-1}+(k-1)}{2k-2}.$$ As the numerator of this fraction is between $2k-2$ and $2k-1$ its integer part is $1$. Now consider $$x_2=\frac1{x_1-1}=\frac{2k-2}{\sqrt{k^2-1}-(k-1)}=\sqrt{k^2-1}+k-1.$$ Its integer part is $2k-2$. Then $$x_3=\frac1{x_2-(2k-2)}=x_1$$ and we have got into a cycle.