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
 A: 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 $x$ ...
A: Extended HINT:  
You need to do first 3 steps manually and make sure you get correctly the first 3 terms
$n-1, 1, 2n-2$ 
Then a recurring pattern should emerge because there is periodicity there.
I mean, say after getting $n-1$ you're left with a number A (to represent as a continued fraction).
Then after getting $n-1, 1, 2n-2$ you should be left with the same number A to represent.    
And that's all, that should be enough for you to conclude that from there on everything just repeats infinitely with a period of 2.   
Just follow the definition of these continued fractions and how to compute them.
That should suffice, I think.   
