Why are these two series identical? Could someone please show/explain to me explicitly why this is true (from wiki):
$S = 1 - 1 + 1 - 1 + 1 - 1 + \cdots$
The series can be rearranged as:
$S = 1 - (1 - 1 + 1 - 1 + 1 - 1 + \cdots)$
I know this should be obvious, but I'm blaming the flu.
 A: A series is characterised by its sequence of partial sums. The two series have the same sequence of partial sums, if we look at a suitable tail. This means either both converge or both diverge, and if they converge, they do so to the same limit.
The first has $1,1-1,1-1+1,1-1+1-1,\ldots$ for partial sums while the second has $1-1,1-(1-1)=1-1+1,1-(1-1+1)=1-1+1-1,\ldots$.
A: Alex's answer explains why the sums are formally identical. I'd like to explain why such manipulations with divergent series make sense at all. Consider the following sum for $x<1$:
$S(x)=1-x+x^2-x^3+x^4-x^5+...$ 
This is a familiar geometric series that sums up to $1\over{1+x}$. Now take the limit $x\to 1$ and you get your limit value $S(1)=\frac{1}{2}$.
Incidentally, notice that $S(x)=1-xS(x)$ for the same reasons the other's explained. One can derive $S(x)=\frac{1}{1+x}$ from that as well.
A: Multiply by $-1$:
$$
a-b+c=a-(b-c)
$$
A: 1 + 1(-1 + 1 + -1 + 1 + -1...)
= 1 + -1 * (1 + 1 + -1 + 1 + -1...)
negative one times a sum of numbers surround by a parenthesis flips the signs
-(x+y) =-1 * (x+ y) = (-x + -y)
