Geometric series/Real/Convergence/Absolutely/Fact

For all real numbers ${}x$ with ${}\vert {x}\vert <1$, the geometric series ${}\sum _{k=0}^{\infty }x^{k}$ converges absolutely, and the sum equals

${}\sum _{k=0}^{\infty }x^{k}={\frac {1}{1-x}}\,.$

Proof, Write another proof