I've recently run into a proof that claims that $1 - 1 + 1 - 1 + 1 - 1 ... = \frac{1}{2}$ that proceeds as follows:
Let $S = 1 - 1 + 1 - 1 + 1 - 1 + ...$. Then $$S = 1 - (1 - 1 + 1 - 1 + 1 - 1 ...) = 1 - S$$ Therefore, $2S = 1$, so $S = \frac{1}{2}$. QED
I was under the impression that this sum doesn't converge, and therefore this step of the proof is invalid:
Let $S = 1 - 1 + 1 - 1 + 1 - 1 + ...$
That is, it's not even valid to suppose that the sum is equal to some (implicitly real or complex) number $S$, and therefore the reasoning that follows is meaningless because $S$ doesn't exist and therefore has any properties we want it to have.
Is the reasoning I've given above correct? That is, is it correct for me to claim that the proof fails because $S$ doesn't exist?
(I've heard that there are techniques for evaluating divergent integrals by using complex analysis and Taylor series, so perhaps there's another way to prove that the summation is $\frac{1}{2}$; I just wanted to see whether my reasoning is sufficient to explain why this particular proof is incorrect. If the proof actually is correct, then I stand corrected!)
Thanks!