Put the continued fraction equal to x, and make the substitution further into the fraction to get
$$x=\frac{1}{1+\frac{1}{2+x}}}$$
Tidying that up gets you quadratic $$x^{2}+2x-2=0,$$ from which $$x=\sqrt{3}-1.$$
thanks Bertie,
I did not have a clue what to make of this question.
you got a mention in my NewYear 2015 post
Bertie....let me ask you a question about this one.....when I carry it out for a few iterations, I get that the results seem to "oscillate' between both roots of that quadratic, i.e., sqrt(3)±1
Did I do something wrong???
Successive iterates are 1, 2/3, 3/4, 8/11, 11/15, 30/41, 41/56, 112/153, 153/209, 418/571, 571/780 ...
They are converging to sqrt(3) - 1 = 0.7320508... .
The last one that I've listed, 571/780 = 0.732051 6dp.
CPhill just sent me this on continued fractions for anyone who is interested :)
http://webserv.jcu.edu/math//vignettes/continued.htm
Thanks CPhill