Kabbalistic Fraction

First we evaluate $x = 2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2 + \frac{1}{2+1} ...}}}$

$\rightarrow x = 2 + \frac{1}{x}$

$x^2 = 2x + 1$

$x^2 - 2x -1 = 0$

$(x-1)^2 - 2 = 0$

$(x- 1- \sqrt{2})(x - 1 + \sqrt{2}) = 0$

$x = 1 + \sqrt{2}, 1 - \sqrt{2}$

As $1 < \sqrt{2}$ , this implies that the second root is negative and thus is not a solution.

Therefore $x = 1 + \sqrt{2}$

The question is asking for $x - 1$ , which is equal to $1 + \sqrt{2} - 1 = \boxed{\sqrt{2}}$

The answer is easy to find by looking the options carefully. Since this sum is like $1 + ......$ So it is greater than 1. So $\frac{1}{2}$ and $e^i\pi$ . Further, this sum is obviously less than 2 because 1 upon a number greater than 1 is always less than 1. So $e$ is wrong............... Hence $\sqrt{2}$ is the correct answer.

y-1=1/(2+(y-1)) Game over.