First, set this equal to x:

\(x = \frac{1}{1+\frac{1}{2+\frac{1}{1+\frac{1}{2+...}}}}\)

We can replace the bottom part with x, shown below:

\(x = \frac{1}{1+\frac{1}{2+x}}\)

Now, solve for x:

\(x = \frac{1}{\frac{2+x+1}{2+x}}\\ x = \frac{1}{\frac{3+x}{2+x}}\\ x=\frac{2+x}{3+x}\\ x^2+3x=2+x\\ x^2+2x-2=0 \)

After using the quadratic formula or completing the square, we find that \(x = -1-\sqrt{3}\) or \(x = \sqrt{3}-1\). Since x must be positive, the negative solution is extraneous, and therefore the answer is \(\boxed{\sqrt{3}-1}\)

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