ExampleConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passverified 2026-08-10 (gpt-5.6-terra-codex-subscription)

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The geometric series represents $1/(1-x)$ for $|x|<1$ and re-expands explicitly about every $c$ with $|c|<1$

Statement

For $|x|<1$ ,

$\frac1{1-x}=\sum_{n=0}^{\infty}x^n.$

More generally, if $|c|<1$ , then

$\frac1{1-x}=\sum_{n=0}^{\infty}\frac{(x-c)^n}{(1-c)^{n+1}}\qquad(|x-c|<1-c).$

Facts & Assumptions

Given: A real $c$ with $|c|<1$ .

[L1]

The geometric series sums to $1/(1-t)$ for $|t|<1$ (For $|r| < 1$ , $\sum_{k \ge 0} r^k = 1/(1-r)$ , and for $|r| \ge 1$ the series diverges).

[L2]

Power-series sums re-expand about interior points, and a nonzero local denominator has a reciprocal series (A power-series sum may be re-expanded about every interior point, with coefficients given by its derivatives there, A convergent real power series with nonzero constant term has a convergent reciprocal power series on a smaller neighbourhood).

Verification

technique · direct

1.1

Apply [L1] with $t=x$ to get the first formula.

2.1

The general results in [L2] show qualitatively that the sum re-expands about $c$ and that the nonzero denominator there has a local reciprocal series. To identify that series and its full convergence interval directly, use $1-x=(1-c)(1-(x-c)/(1-c))$ and $1-c>0$ , and apply [L1] with $t=(x-c)/(1-c)$ . This gives the second formula precisely when $|x-c|<1-c$ .

givenL1L2algebra∎

Depends on

Used by

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Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 101 results over 23 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click the chart to enlarge it.

Sources

Power series, Encyclopedia of Mathematics (standard reference, not scraped)
Northwestern Math 320-2 lecture notes (standard reference, not scraped)

The geometric series represents 1/(1−x)1/(1-x)1/(1−x) for ∣x∣<1|x|<1∣x∣<1 and re-expands explicitly about every ccc with ∣c∣<1|c|<1∣c∣<1