Theorem T35 If 0≤x<10 \leq x < 10≤x<1, then ∑n=0∞xn=11−x .\sum_{n=0}^\infty x^n = \frac{1}{1 - x} \; . n=0∑∞xn=1−x1. If x≥1x \geq 1x≥1, the series diverges.