TheoremStatement: AI-adaptedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passaudited 2026-07-31

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A power series converges absolutely and uniformly on every closed interval strictly inside its interval of convergence

Statement

Let $\sum a_n(x-c)^n$ have radius $R$ , and let $[u,v]$ be a nonempty closed interval for which

$r:=\max\{|u-c|,|v-c|\}<R.$

Then the function series converges absolutely at every point of $[u,v]$ and converges uniformly there.

Facts & Assumptions

Given: A power series of radius $R$ and a closed interval $[u,v]$ satisfying the strict interior condition above (Intervals of $\mathbb{R}$ : the nine order-convex forms, nondegeneracy, and length, A series of real-valued functions and its pointwise and uniform convergence through its partial sums).

[L1]

The power series converges absolutely at every point whose distance from $c$ is less than $R$ (A real power series converges absolutely inside its radius and diverges outside it, while either behaviour may occur at an endpoint).

[L2]

If $|f_n(x)|\le M_n$ for all $n,x$ and $\sum M_n$ converges, the Weierstrass M-test gives absolute pointwise and uniform convergence of $\sum f_n$ (The Weierstrass M-test gives absolute pointwise convergence and uniform convergence of a function series).

Proof

technique · direct

1.1

Choose a real $\rho$ with $r<\rho<R$ , or merely $\rho>r$ when $R=+\infty$ . Then the scalar series $\sum |a_n|\rho^n$ converges by [L1], applied at $x=c+\rho$ .

givenL1choose

1.2

For every $x\in[u,v]$ , order-convexity gives $|x-c|\le\max\{|u-c|,|v-c|\}=r<\rho$ , and hence $|a_n(x-c)^n|\le |a_n|\rho^n$ for every $n$ .

givenalgebra

2.1

Apply [L2] to $f_n(x)=a_n(x-c)^n$ and $M_n=|a_n|\rho^n$ . The series is absolutely convergent at each $x\in[u,v]$ and uniformly convergent on the whole interval.

step 1.1step 1.2L2∎

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 68 results over 14 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

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