Geometric series as a power series

Example

Let $R$ be a commutative ring and let $a \in R$. In the ring $R[\![X]\!]$ of formal power series, $$(1 - aX) \sum_{n \ge 0} a^n X^n = 1.$$ Thus $1 - aX$ is a unit, and $$\frac{1}{1 - aX} = \sum_{n \ge 0} a^n X^n.$$

Remarks

The proof is a telescoping product. The coefficient of $X^n$ in the product is $a^n - a \cdot a^{n-1} = 0$ for $n \ge 1$, and it is $1$ for $n = 0$. With $a = 1$, it gives $\frac{1}{1 - X} = 1 + X + X^2 + \dots$, the generating function of the constant sequence $1$. Over $\mathbb{R}$ the same identity holds as a convergent series for $|aX| < 1$, but the formal identity needs no convergence.

Explanations

L1 human

Multiply $1 + aX + a^2 X^2 + \dots$ by $1 - aX$. Every term cancels with the next one, and only $1$ is left. So the geometric series is the inverse of $1 - aX$.