Continued fractions paradox
When we evaluate a continued fraction, we're trying to solve an equation: in this case, $$x={2\over 3-x}.$$
This turns into $3x-x^2=2$, or $x^2-3x+2=0$; or, better yet, $(x-2)(x-1)=0$. This has two solutions, namely $x=2$ and $x=1$.
The "contradiction" here comes from the implicit assumption that a continued fraction always describes a single number - in the same way that a decimal expansion picks out a specific real - as opposed to being a description that could apply to several things. Basically, the point is that $$x={2\over 3-{2\over 3-{2\over 3-...}}}$$ is really just a tricky way of saying $$\mbox{"$x$ has the property that $x^2-3x+2=0$,"}$$ which of course doesn't refer uniquely. The expression $${2\over 3-{2\over 3-{2\over 3-...}}}$$ looks like something that refers to a specific number, but actually doesn't.
Fundamentally, this is an error where you see something that "looks like" a calculation, and you assume it actually means something. But it doesn't. The expression:
$$\dfrac{2}{3-\dfrac{2}{3-\frac{2}{3-\dots}}}$$ looks like something, but it isn't well defined. Is it the limit of:
$$\frac{2}{3},\dfrac{2}{3-\frac{2}{3}},\dfrac{2}{3-\dfrac{2}{3-\frac{2}{3}}},\dots$$
or is it the limit of:
$$2,\dfrac{2}{3-2},\dfrac{2}{3-\frac{2}{3-2}},\dots?$$
The first sequence converges to $1$, the second sequence converges trivially to $2$ (since every value is $2$.)
It's worth noting that if $x_0$ is defined as any value, and $x_{n+1}=\frac{2}{3-x_n}$, this sequence actually only converges to $2$ if $x_0=2$. This is because:
$$2-\frac{2}{3-x}=\frac{4-2x}{3-x}=\frac{2}{3-x}(2-x)$$
so $|x_{n+1}-2|>|x_n-2|$ when when $x_n>1$ and $x_n\neq 2$. So the only way for it to converge to $2$ is if it is always equal to $2$.
You get this problem, too, if you just take a simple equation:
$$x = 1+\frac{1}{1+x}$$
which has solutions $\pm \sqrt{2}$. That seems to mean that the continued fraction:
$$1+\frac{1}{2+\frac{1}{2+\dots}}$$
is likewise ambiguous. The only reason we don't see it as a paradox is that intuitively, we know this is only the positive value.
So the real problem is in extrapolating from the equation $\gamma = a+\frac{b}{c+\gamma}$ to some infinite form, which may ore may not have meaning.
A similar error can be seen more simply if you start with:
$$2=1+(-1)+2$$
You would never conclude that $$2=1+(-1)+1+(-1)+\cdots$$
The finite sums always equal $2$, but the $2$ is no longer present in the infinite sum, and you the meaning you ascribe to the infinite sum depends on definitions.
If you look very carefully at each and every step, the "paradox" ends up being a "sleight of hand". When it says "I'll use the ellipsis", it means to indicate that any number of finite steps can be performed. The second time it does not even say "ellipsis", but it means the same. But then, at the end, it says "this is exa[c]tly the same continued fraction". It is not. They are not continued fractions: they are expressions with finite steps. Continued fractions (as you can see in the other answers) are infinite expressions with some careful definition of convergence.
(The sleight of hand is that they look the same as the usual form for continued fractions, by using a meaning for ellipsis that is different from the meaning it has in continued fractions.)
[You should really make your question stand alone, by the way.]