Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passaudited 2026-08-02
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The power series for log(1+x) on (-1,1], including the Abel endpoint

Statement

For 1<x1-1<x\le1, log(1+x)=n=1(1)n+1xnn.\log(1+x)=\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}. The series converges at x=1x=1 to log2\log2 and diverges at x=1x=-1.

Facts & Assumptions

Given: A real xx with 1<x<1-1<x<1.

[L1]

log(u)=1/u\log'(u)=1/u on (0,)(0,\infty), so (log(1+x))=1/(1+x)(\log(1+x))'=1/(1+x) and log1=0\log1=0 (The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t).

[L2]
[L3]

A real power series may be integrated term by term inside its radius of convergence (Inside its radius a real power series may be integrated term by term on every closed subinterval).

Proof

technique · direct
1.1

Integrating the series of [L2] from 00 to xx gives n1(1)n+1xn/n\sum_{n\ge1}(-1)^{n+1}x^n/n.

L2L3
2.1

By [L1], the integral of 1/(1+t)1/(1+t) from 00 to xx is log(1+x)\log(1+x), so the displayed series formula holds for 1<x<1-1<x<1.

L1step 1.1
3.1

At x=1x=1 the series is alternating harmonic and converges by [L4]; Abel's theorem and step 2.1 identify its sum with log2\log2.

step 2.1L4
4.1

At x=1x=-1 every term is 1/n-1/n, so the series is the negative harmonic series and diverges.

L4algebra

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Direct dependencies and their dependencies through the next three levels: 148 results over 27 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.

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