How to show that $ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $? By long division, it is easy to show that $$ \frac{1}{1-x} = 1 + x + x^2 +x^3 +... $$
But how to show that
$$ 1 + x + x^2 +x^3 +...= \frac{1}{1-x} $$

Suppose $1+x+x^2+x^3...=S$ where $S \in R$.
If $x=0$ it is trivial, so suppose $x \neq 0$. Subtract each side by 1 and divide both sides by $x$.
This leaves: $S=(S-1)x^{-1}$. Solving for $x$ yields: $S*x=S-1$ which simplifies to $S(x-1)=-1$ which then yields $1+x+x^2+x^3...=(1-x)^{-1}$
Note domain restrictions of $|x|<1$ applies.
