<s>[INST] $1=2$ | 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 $1=2$. 
 [/INST] This is a general feature of continued fractions.
What you are doing is iterating the function 
$$ s(x) = \frac{1}{2-x} $$
... so, for instance, you have that
$$ 1 = s(1) = s(s(1)) = s(s(s(1))) = \cdots $$
and
$$ 2 = s(3/2) = s(s(4/3)) = s(s(s(5/4))) = \cdots .$$
The function $s$ has a single fixed point at 1 (first line).  It is known that iterating $s$ from any starting point except $x=2$ converges to 1 (think of $s$ as a mapping of the Riemann sphere; it takes $2 \mapsto \infty$ and $\infty \mapsto 0$). For $x<1$, this is easy to check: $x<s(x)<1$.
So, if $x\neq 2$, then
$$ \lim_{n \to \infty} s^n(x) = 1 , $$
where $s^n(x)=s(s(\cdots s(x)))$ is the $n$th iterate of $s$.
But if you start with some $y$, it's always possible to find an $x_n$ so that $s^n(x_n) = y$ (including $\infty$, the map is one-to-one).  You've shown that $$s^n\left(\frac{n+2}{n+1}\right) = 2.$$
Another nice thing to note here is that to end up with a limit above 1, we need  the numbers at the end of the fractions (the $x_n$) to approach 1 from above: i.e., if for all $n$ that
$$ s^n(x_n) = y > 1  $$
then
$$ x_n \searrow 1 .$$
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