The geometric series $\sum x^k$ converges exactly for $\abs{x}<1$, with sum $1/(1-x)$.Suggest a slogan
Let \(x\in\CC\). The series \(\sum_{k=0}^\infty x^k\) converges if and only if \(\abs{x} < 1\), and in that case \[ \sum_{k=0}^\infty x^k = \frac 1 {1-x} . \]