Self Referencing Answer

If $1 - x + x^2 - x^3 +...$ is the answer to this question , then $1 + x + x^2 + x^3 +... = 1 - x + x^2 - x^3 +...$ will have to be true .We have two possiblities:

$\boxed{1}$ .- $|x| \ge 1$ . In this case, $1 + x + x^2 + x^3$ or $1 - x + x^2 - ...$ doesn`t converge. And if one of them converged, the another one wouldn't converge and then $1 + x + x^2 + x^3 +... \neq 1 - x + x^2 - x^3 +...$

$\boxed{2}$ .- $|x| < 1$ then $1 + x + x^2 + x^3...$ and $1 - x + x^2 - x^3 +...$ are absolutely convergent and hence they are inconditionally convergent and therefore their sum or substraction also are. This means we can order the terms in every ways and since $1 + x + x^2 + x^3 +... = 1 - x + x^2 - x^3 +...$ $\Rightarrow 0 = 2(x + x^3 + x^5 +...)$ $\Rightarrow x = 0$ (because if $x > 0$ then $S = x + x^3 + x^5$ would be $S > 0$ and if $x < 0$ would happen $S < 0$ ) and hence $1 - x + x^2 - x^3 +... = 1 + x + x^2 +... = 1$ and one solution exists being this one.

$\boxed{2}$ .- $|x| < 1$ then $\displaystyle \frac{1}{1 - x} = \sum_{n = 0}^\infty x^i = 1 + x + x^2 +....$ and $\displaystyle \frac{1}{1 + x} = \sum_{n = 0}^\infty (-1)^n \cdot x^i = 1 - x + x^2 - x^3 +....$ . Since, it has to be $\frac{1}{1 - x} = \frac{1}{1 + x} \Rightarrow x = 0$ ...

P-S.- I'm talking in $\mathbb{R}$