Grandi's series contradiction This is the Grandi's series: $1-1+1-1+1-1+\dots$

The series can be equal to $0$ 

$$(1-1)+(1-1)+(1-1)+\dots=0+0+0+\dots=0,$$

or to $1$ 

$$1-(1-1)-(1-1)-(1-1)-\dots=1-0-0-0\dots=1,$$

or to $1/2$ 

$$S=1-1+1-1+1-\dots,\quad\quad    S=1-(1-1+1-1+1-1+1-...)$$
$$\Rightarrow   S=1-S\Rightarrow     2S=1\Rightarrow    S=1/2$$

Isn't this a contradiction? The integers are closed under addition and subtraction, but we get a fraction. Why?
The issue here is that addition of a finite number of terms is a nice, well-defined operation; addition of an infinite number of terms need not be.

In particular, rearranging terms in an infinite sum can actually change the sum. The series you have presented does not converge - and so discussing it as having a value is incorrect to begin with.  However, there are series which converge, and which STILL cannot be rearranged without changing the value: this is true for series that are called conditionally convergent - that is, the series converges, but the sum of absolute values of the terms does not.

Interestingly, for conditionally convergent series, it turns out that you can rearrange the terms to give any limit that you like.

At the end of the day, the big thing to remember is this:  addition of an infinite number of terms cannot be assumed to be associative or commutative, unless you know that the sum of the absolute values of the terms converges - a property known as absolute convergence.
$$S(5) = 1-1+1-1+1 = 1$$
then   $1-S(5) = 1-1+1-1 = S(4)$.  So  $1-S(n) = S(n-1)$.  If $S(5)=S(4)$ (which is not), then $S(5)=1/2$, wrong, and similar to $S=1/2$. Grandi assumes  $S(\infty)=S(\infty-1)=S$, so that $S=1/2$.

The question is: Is it really $S(\infty)=S(\infty-1)$?. Of course $\infty-1= \infty$, but this is not the case, now $\infty$ is not a value, is the order of the variable which still will take values of $1$ or $-1$, even their order is $\infty$.

So $S(\infty)$ does not equal $S(\infty-1)$. That's what I believe.