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Converging series

Fri, Apr 21, 2017 02:00 CEST

Tags:Identities

Let $$-1 < x < 1$$,

$$1 + x + x^2 + x^3 + x^4 + \cdots = \frac{1}{1-x}.$$

Proof

\begin{aligned} (1 - x)(1 + x + x^2 + x^3 + x^4 + \cdots + x^k) &= 1 - x^{k+1}, \\ \lim_{k \to \infty} (1 - x)(1 + x + x^2 + x^3 + x^4 + \cdots + x^k) &= \lim_{k \to \infty} 1 - x^{k+1}, \\ (1 - x)(1 + x + x^2 + x^3 + x^4 + \cdots) &= 1. \end{aligned}

A fun way to prove it for $$x = \frac1b$$ with $$b \in \mathbb{N}_0$$ is to note that for $$a = b-1$$

$$1.111\ldots_b = \frac{a.aaa\ldots_b}{a} = \frac{10_b}{a} = \frac{b}{a} = \frac{b}{b-1} = \frac{1}{1-x}.$$