Paper folding and continued fractions This question is suggested by a prior question, which has not received a complete answer.  One corner of rectangular sheet of paper $P_1$ is folded down to make a trapezoid, and then a right triangle is cut off to make another rectangle $P_2.$  The process is repeated with $P_2$ to produce a rectangle $P_3$ and so on.  We assume that the ratio of the sides of $P_1$ is irrational, so none of the rectangles is a square, and the process goes on forever.

Let $S_n$ be the area of $P_n$ and suppose $$\lim_{n\to\infty}{S_{n+2}\over S_n}$$ exists.  What are the possible values of the limit.  
In the original question, the OP stated that he believed that, in order for the limit to exist, it must be the case that $P_n$ and $P_{n+2}$ are similar, for large $n,$ but that he couldn't prove it. 
In a partial answer, I corrected some calculation errors, and pointed out the relation of the problem to continued fractions, but I haven't been able to prove the OP's hypothesis either, though I think it's very likely true.
Can you prove that $P_n$ and $P_{n+2}$ must eventually be similar?  Alternatively, can you see how to modify the analysis so as not to use this hypothesis?

There is another possible limit $\color{blue}{3-2\sqrt{2}=(1+\sqrt{2})^{-2}}$.  How?
Suppose rectangle $n$ has length $a$ and width $b$ with $a>2b$.  Then to get to rectangle $n+2$ you shorten the $a$ dimension twice leaving a rectangle with dimensions $a-2b$ and $b$.  If
$\dfrac{a-2b}{b}=\dfrac{b}{a}$
then the rectangles are similar forcing a fixed value of $S_{n+2}/S_n$.  Note that the intermediate rectangle would not be similar so we would not have $S_{n+1}/S_n=\sqrt{S_{n+2}/S_n}$.
The above equation for $a$ and $b$ is solved for a positive root $a/b=1+\sqrt{2}>2$.  Thereby
$\dfrac{S_{n+2}}{S_n}=\dfrac{a-2b}{a}=3-2\sqrt{2}$.
With $a/b=1+\sqrt{2}>2$ one can verify that also 
$\dfrac{S_{n+3}}{S_{n+1}}=3-2\sqrt{2}$.
