Alphabeta Math
ExampleConstruction: AI-generatedVerification: AI-generatedprecheck passaudited 2026-08-11
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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

  • Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
  • AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
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A step function whose improper integral is the alternating harmonic series

Example

For each nonnegative integer k, define f(x)=(−1)kk+1(k≤x<k+1). Then ∫0∞f converges conditionally, and its value is the sum of the alternating harmonic series.

Facts & Assumptions

Verification

technique · computation
1.1

Every compact interval meets only finitely many jumps, so [L3] makes f properly integrable there. At a positive integer N, [L1, L3] ∫0Nf=∑k=0N−1(−1)kk+1, which converges as N→∞ by [L1].

1.2

If N≤R<N+1, the remaining integral from N to R has absolute value at most 1/(N+1). Hence arbitrary real truncations have the same limit as the integer truncations.

given
2.1

At integer truncations, ∫0N∣f∣=∑k=0N−11/(k+1), unbounded by [L2]. Thus the integral converges but not absolutely.

L2∎

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

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Sources