Geometric series/Real/Convergence/Absolutely/Fact/Proof

Proof

For every ${}x$ and every ${}n\in \mathbb {N}$ we have the relation

${}(x-1){\left(\sum _{k=0}^{n}x^{k}\right)}=x^{n+1}-1\,$

and hence for the partial sums the relation (for ${}x\neq 1$ )

${}s_{n}=\sum _{k=0}^{n}x^{k}={\frac {x^{n+1}-1}{x-1}}\,$

holds. For ${}n\rightarrow \infty$ and ${}\vert {x}\vert <1$ this converges to ${}{\frac {-1}{x-1}}={\frac {1}{1-x}}$ because of fact and exercise.

To fact