$1=2$ | Continued fraction fallacy It's easy to check that for any natural $n$
$$\frac{n+1}{n}=\cfrac{1}{2-\cfrac{n+2}{n+1}}.$$
Now, 
$$1=\frac{1}{2-1}=\frac{1}{2-\cfrac{1}{2-1}}=\frac{1}{2-\cfrac{1}{2-\cfrac{1}{2-1}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-1}}}}=\ldots
=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\dots}}}}},$$
$$2=\cfrac{1}{2-\cfrac{3}{2}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{4}{3}}}=\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{5}{4}}}}=\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\frac{1}{2-\frac{6}{5}}}}}=\ldots
=\cfrac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\frac{1}{2-\ldots}}}}}.$$
Since the right hand sides are the same, hence $1=2$. 

Incidentally, nobody appeared to have resolved the fallacy of the question, so I have provided an answer.

For all $a\in \mathbf N$, it follows $$\cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots - \cfrac 12}}}$$ such that the number of times the reciprocal in the continued fraction appears is $a$.
Proof. Note the identity $$\cfrac{1}{1+a}=1-\cfrac{1}{1+\color{red}{\cfrac 1a}}.$$ By letting $a=b-1$, it follows $$\cfrac 1b = 1-\cfrac{1}{1+\cfrac{1}{b-1}}.$$ From this we can substitute for $\color{red}{\cfrac 1a}$. $$\therefore \cfrac{1}{1+a}=1-\cfrac{1}{2-\cfrac{1}{1+\cfrac{1}{a-1}}}.$$ Clearly we can now likewise substitute for $1/(a-1)$, and the pattern will continue until for some $k\in\mathbf N$, the denominator of $1/(a-k)$ reaches $a-k=1$ since it cannot pass $0$. In consequence, we deduce as desired. (And, of course, when $a=0$, we have $1/1 = 1-0$.) This completes the proof. $\;\bigcirc$
And now, since $$\lim_{a\to\infty}\frac{1}{1+a}=0$$ then $$\boxed{\cfrac{1}{2-\cfrac{1}{2-\cfrac{1}{2-\ddots}}}=1}$$
