CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

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Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
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The geometric series converges pointwise but not uniformly on $(- 1, 1)$

Statement

The partial sums of $\sum_{n \geq 0} x^{n}$ converge pointwise to $1 / (1 - x)$ on $(- 1, 1)$ but do not converge uniformly there.

Facts & Assumptions

Given: The geometric-series partial sums on $(- 1, 1)$ .

[L1]

Pointwise convergence follows from the geometric-series theorem (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

Uniform convergence implies the uniform Cauchy property (Pointwise convergence, uniform convergence, and the uniformly Cauchy condition for sequences of real-valued functions).

Verification

technique · direct

1.1

Consecutive partial sums differ by $x^{N}$ , whose supremum over $(- 1, 1)$ is $1$ for every $N$ .

givenalgebra

2.1

Thus the partial sums are not uniformly Cauchy by [L2] and hence not uniformly convergent, though [L1] gives pointwise convergence. This is the counterexample recorded in FALSE: every power series converges uniformly on its entire open interval of convergence.

step 1.1L1L2∎

Depends on

Used by

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Dependency tree · two levels

17 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

MIT 18.100C, Lecture 11: Power Series (standard reference, not scraped)
Power series, Encyclopedia of Mathematics (standard reference, not scraped)

The geometric series converges pointwise but not uniformly on (−1,1)(−1,1)