Is the series $ 1-1+1-1+1+\cdots = -1+1-1+1-1+1-\cdots $? We have the series
$$ S = 1-1+1-1+1-1+1 \cdots $$
If we manipulate S, we get that
$$ 1 - S = S $$
$$ 1 = 2S $$
$$ \frac{1}{2} = S $$
Also, if we re-order S, we get that
$$ S = (1-1)+(1-1)+(1-1)\cdots $$
$$ = -1+1-1+1-1+1-1+1\cdots $$
So $$ -S = S + 1  $$
Solving for S we get that
$$ S = -\frac{1}{2} $$
My question is, is the new S re-ordered considered the same series as S? When we can say that two series are the same in that context? Thanks.

One of the simplest way to get this series will be with the help of Binomial Expansion of $\frac{1}{2}$ as follows
$\frac{1}{2} = \frac{1}{1+1 }= (1+1)^{-1}$
Expansion will be as follows
$1^{-1} + \frac{(-1)}{1!}(1)^{-2} \cdot(1)^1+\frac{(-1)\cdot(-2)}{2!}1^{-3}\cdot1^2+...  = 1-1+1-1+…$
Clearly now in your question
LHS = $\frac{1}{2} $
RHS = $-1+\frac{1}{2} =  \frac{-1}{2} $
In this context RHS $\neq$ LHS
Some more interesting series with Binomial Expansion will be awesome, may be sometimes divergent series also
For example
Expansion of $\frac{1}{3}$ can be done in 2 ways
Conventional and logic way of expansion will be $$\frac{1}{3} = \frac{1}{2+1 }= (2+1)^{-1}$$ which will expand into
$$ 2^{-1}+ \frac{-1}{1!}\cdot 2^{-2}+\frac{{(-1)}\cdot{(-2)}}{2!}\cdot2^{-3}+\frac{{(-1)}\cdot{(-2)}\cdot(-3)}{3!}\cdot2^{-4}...$$
$$=\frac{1}{2}-\frac{1}{4}+\frac{1}{8}-\frac{1}{16}+...$$
Another way to put this Binomial Expansion will be $$\frac{1}{3} = \frac{1}{1+2}= (1+2)^{-1}$$
Expansion now will give different result
$$ 1^{-1}+ \frac{-1}{1!}\cdot 1^{-2}\cdot2^1+\frac{{(-1)}\cdot{(-2)}}{2!}\cdot2^{2}+\frac{{(-1)}\cdot{(-2)}\cdot(-3)}{3!}\cdot2^{3}...$$
And this will be $$1-2^1+2^2-2^3+...$$ a divergent series alternating from $-\infty$ to $+\infty$
