Prove: $\sqrt{(n^{2}-1)}= [n-1,1,2n-2,1,2n-2,1,2n-2, \ldots ]$ I'm trying to show 
$$
\sqrt{n^2-1}=n-1 + \cfrac{1}{1 + \cfrac{1}{2n-2+ \cfrac{1}{1+\cdots}}}
$$
What I tried:
$$
\sqrt{n^2 - 1}- (n-1) = \frac{2n-2}{n^2-1+(n-1)}
$$
And here i got stuck on how to continue, would greatly appreciate any help

You want to show that
$n-1+\sqrt{n^2-1}=2n-2 + \cfrac{1}{1 + \cfrac{1}{2n-2+ \cfrac{1}{1+\cfrac{1}{2n-2+\cdots}}}}$
Suppose
$y=x+\cfrac{1}{1+\cfrac{1}{x+\cfrac{1}{1+\cfrac{1}{x+\cdots}}}}$
then
$y = x + \cfrac{ 1}{1+\frac 1 y} \\\Rightarrow y=x+\frac {y}{y+1}\\\Rightarrow y^2+y = xy+x+y\\\Rightarrow y^2 - xy - x = 0 \\\Rightarrow y=\frac{x \pm \sqrt{x^2+4x}}{2}$
Now substitute $2n-2$ for $x$ ...
