False statementConstruction: AI-adaptedVerification: AI-generatedprecheck passaudited 2026-07-31

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FALSE: every power series converges uniformly on its entire open interval of convergence

Statement

False claim: every real power series converges uniformly on its entire open interval of convergence.

Facts & Assumptions

Given: The geometric power series $\sum_{n \geq 0} x^{n}$ on $(- 1, 1)$ .

[L1]

It converges pointwise there to $1 / (1 - x)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[L2]

A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy (A sequence of real-valued functions converges uniformly if and only if it is uniformly Cauchy).

Refutation

technique · direct

1.1

For every $N$ , the difference between the $(N + 1)$ st and $N$ th partial sums is $x^{N}$ . Its supremum over $x \in (- 1, 1)$ is $1$ .

givenalgebra

2.1

Thus the partial sums are not uniformly Cauchy and cannot converge uniformly by [L2], despite pointwise convergence on the entire open radius interval by [L1].

step 1.1L1L2∎

Depends on

Used by

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

Dependency tree · two levels

18 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)