Checking my understanding: $1 - 1 + 1 - 1 + 1 - ... = \frac{1}{2}$ 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!
 A: If you have some method of attaching a sum to a series (e.g. Cesaro summation, as discussed in Ayesha's answer, or Abel summation) which is linear (i.e. the limit of a linear combination of two series coincides with the same linear combination of the limits), and if $1- 1 + 1 - 1 + \cdots$ is summable with respect to this method, then this argument shows that the value of the sum will
have to be $1/2$.
By itself, this argument won't tell you whether a given summation method applies
to your series, though.  
A: The series of course doesn't converge - that doesn't mean it doesn't have a sum. Indeed, note that the Cesaro means of the series tend to 1/2, and thus the Cesaro sum is 1/2.
Consider the partial sums of the series $s_n$. Then we say the (C, 1) sum of the series is the limit $$\lim_{n \to \infty} \frac{1}{n}\sum_{k=1}^{n}s_k $$ if this limit exists. 
For your series here, we have the partial sums $1, 0, 1. . .$ and thus the Cesaro means are $1, \frac{1}{2}, \frac{2}{3}. . . $ a sequence tends to 1/2. Your proof is a heuristic demonstration of this rigorous fact.
