Is there anything in the proof for $\frac{a}{1-r}$ to show that it won't work for $|r|\gt1$ when summing an infinite geometric series? Why doesn't $\frac{a}{1-r}$ work for $|r|\gt1$ when summing an infinite geometric series?

The proof is pretty short:
$$S= a + ra + r^2a + r^3a + ...$$
$$rS = ra + r^2a + r^3a + ... = S - a$$
$$a = S - rS = S(1 - r)$$
$$S =\frac{a}{1-r}$$

It's clear from inspection that this is undefined for $r=1$, but the function is defined for all other real $r$. If I plug $r=2$ into the first equation (the infinite sum), I get:
$$S= a + 2a + 2^2a + 2^3a + ...$$
which plainly blows up. Yet, when I plug $r=2$ into the last equation:
$$S=\frac{a}{1-2}= -a$$
Is there anything in the "pretty short proof" that should have alerted me to the fact the the result would only work for $|r|\lt1$?
 A: The problem with the proof is that it assumes that $S$ is a real number (i.e. the series converges) so that manipulations can be done. For example $S-rs$ is meaningless if $S=\infty$. This argument works only if you show that $S$ is a real number apriori. To see why the geometric series doesn't converge when $|r|\geq 1$, recall that in general 
$$
S=\sum_{n=0}^\infty a_n\stackrel{\text{def}}{=}\lim_{n\to\infty}S_n.
$$ 
if the limit exists where $S_n=\sum_{k=0}^na_k$ is the nth partial sum. If the sum converges (i.e. the limit exists), then $a_n\to 0$. Indeed,
$$
\lim_{n\to\infty}a_n=\lim_{n\to\infty}(S_{n+1}-S_n)=\lim_{n\to\infty}S_{n+1}-
\lim_{n\to\infty}S_{n}=S-S=0.
$$
In the case of the geometric series if $|r|\geq 1$, then
$$
|r^n|=|r|^n\geq1
$$
hence $r^n\not \to 0$ and hence the series fails to converge in this case. 
A: Working with the numbers$$S=a+ra+r^2a+\cdots\text{ and }rS=ra+r^2a+r^3a+\cdots$$assumes that these numbers exist, that is, the both series converge. And this happens only when $|r|<1$.
A: The algebraic operations you apply to $S$ only work if $S$ is a finite number;  but (abusing notation) $\infty - r \infty$ is an indeterminate form, and you can't do algebra with it.   What you've proven is that if $S$ is a finite number, then it is equal to $a/(1-r)$.
A: You are working with a sum that is divergent, i.e. goes to infinity. Hence more or less you are having the equation "infinity= infinity" and then substract infinity from one side to get $a=S(1-r)$. Those sums require more care and one way to do that is to ensure the sum is finite by taking $|r|<1$
A: $\displaystyle S= a + ra + r^2a + r^3a + ...=a\sum_{n=0}^{\infty} r^n$
For convergence of this sequence, we should prove:
$\sqrt[n]{\big|r^n|}<1 \to |r|<1$
Otherwise, it diverges.
