I can't seem to figure out why the proof is false. Everyone knows that $1 \ne 2$ and that $1=2$ is only proven through mathematical fallacies.
However I can't figure out where the problem is in the following fallacy:
Observe that
$$1=\frac{2}{3-1}$$
Substitute $\frac{2}{3-1}$ for the $1$ in the denominator
$$1=\frac{2}{3-\frac{2}{3-1}}$$
Substitute again $$1=\frac{2}{3-\frac{2}{3-\frac{2}{3-1}}}$$
Therefore $1$ can be written as the infinite fraction: $$1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$$
Now notice that $$2=\frac{2}{3-2}$$
Substitute 2 similar to how 1 was substituted above
$$2=\frac{2}{3-\frac{2}{3-2}}$$
And through repeated substitution we get $$2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$$
Notice that $1=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$ and $2=\frac{2}{3-\frac{2}{3-\frac{2}{3-\ldots}}}$
Therefore, $1=2$
Just looking at this I know that it cannot be true. But they both seem to share the same infinite fraction, so what is it that I am missing?
If you look at the sequence of convergents for each continued fraction, you'll see they are different sequences. They don't exactly represent series, but you still have to be careful. Just because you can write "dot dot dot" doesn't mean that things really converge absolutely.