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  "body": "\n## Example\n\nLet $R$ be a commutative ring and let $a \\in R$.\nIn the ring $R[\\![X]\\!]$ of [[formal power series|072W67KFYK]],\n$$(1 - aX) \\sum_{n \\ge 0} a^n X^n = 1.$$\nThus $1 - aX$ is a unit, and\n$$\\frac{1}{1 - aX} = \\sum_{n \\ge 0} a^n X^n.$$\n\n## Remarks\n\nThe 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 $n = 0$.\nWith $a = 1$, it gives $\\frac{1}{1 - X} = 1 + X + X^2 + \\dots$, the [[generating function|0HH6JX5WZZ]] of the constant sequence $1$.\nOver $\\mathbb{R}$ the same identity holds as a convergent series for $|aX| \u003c 1$, but the formal identity needs no convergence.\n\n## Explanations\n\n### L1 human\n\nMultiply $1 + aX + a^2 X^2 + \\dots$ by $1 - aX$. Every term cancels with the next one, and only $1$ is left.\nSo the geometric series is the inverse of $1 - aX$.\n",
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