Alphabeta Math
TheoremStatement: AI-adaptedProof: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-29
How statement and proof provenance work

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.
  • AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The integral test: for f≥0 nonincreasing on [0,∞), ∑kf(k) converges if and only if the sequence (∫0Nf)N is bounded, with ∫0Nf≤∑k<Nf(k)≤f(0)+∫0Nf

Statement

Let f:[0,∞)→R be nonincreasing (Nondecreasing, increasing (strictly increasing), nonincreasing, decreasing, monotone and strictly monotone real functions on a subset of R, with the dictionary to monotone sequences) with f(t)≥0 for every t≥0. For N∈N write

JN  :=  ∫0Nf,

which is defined for every N: for N≥1 the restriction of f to [0,N] is monotone, hence bounded and integrable (A monotone function on [a,b] is Riemann integrable: for the uniform partition into N parts the upper minus lower sum telescopes to ∣f(b)−f(a)∣ (b−a)/ι(N)), and J0=∫00f=0 (The integral with oriented limits: ∫aaf:=0 and ∫baf:=−∫abf). Here N inside the integral means the canonical natural ι(N)∈R (The canonical natural ι(n)=n⋅1F of a field), as everywhere in this library. Let sN:=∑k<Nf(k) be the partial sums of ∑kf(k) (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion), the index k ranging over N, which contains 0. Then:

  1. The bracket. For every N∈N, JN  ≤  sN  ≤  f(0)+JN.
  2. The test. (JN)N∈N is nondecreasing, and ∑kf(k) converges if and only if the set { JN:N∈N } is bounded above (Lower bound, bounded below, bounded set).

The conclusion is about a sequence of proper integrals, and that is deliberate. This library has not defined ∫0∞f at this point in the reading order — improper integrals are developed on a later page — so the statement that a reader may expect, "∑kf(k) converges if and only if ∫0∞f converges", is not available and is not made. What is proved is the statement above, which is what that one abbreviates; the later page is where the two are identified.

The index starts at 0. Both the sum and the integral begin at 0, because N contains 0 and a sequence is a function on N (Sequences of reals: bounded, eventually, frequently, tails, subsequences). The classical statement, which starts at 1, is the statement about the first tail of ∑kf(k) and is not the statement above.

Facts & Assumptions

Given: A nonincreasing f:[0,∞)→R with f≥0, and the notation JN=∫0Nf, sN=∑k<Nf(k) for N∈N.

[L4]

Finite sums: telescoping ∑k<n(ck+1−ck)=cn−c0, splitting, additivity, and monotonicity in the terms (Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L5]

For a sequence (ak) of nonnegative reals, the partial sums are nondecreasing and ∑ak converges if and only if the set of partial sums is bounded above (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Series, partial sums, convergence and the sum, divergence, and the tail series, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

ι(0)=0, ι(N+1)=ι(N)+1, and ι is nondecreasing on N (The canonical natural ι(n)=n⋅1F of a field, Canonical naturals are positive and strictly increasing).

[L7]

Ordered-field arithmetic: the order is total and transitive, and adding constants preserves inequalities (Ordered field, Complete ordered field (least-upper-bound property), Intervals of R: the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct
1.1givenL1L6

For k∈N the interval [ι(k),ι(k+1)] is nondegenerate of length 1 by [L6], f is integrable on it by [L1], and f(k+1)≤f(t)≤f(k) for every t in it, f being nonincreasing.

1.2L1L3L4L6

By [L3] and [L4], JN=∫0Nf=∑k<N∫kk+1f for every N∈N: writing Φ(x):=∫0xf, each summand is Φ(k+1)−Φ(k) and the sum telescopes to Φ(N)−Φ(0)=JN.

1.3L4

By [L4], ∑k<Nf(k+1)=sN+1−f(0), since splitting sN+1=∑k<N+1f(k) at index 1 gives f(0)+∑j<Nf(1+j).

2.1step 1.1L2

Hence f(k+1)≤∫kk+1f≤f(k) for every k∈N, by [L2] with q−p=1.

3.1step 2.1step 1.2L4

Summing step 2.1 over k<N with [L4] gives ∑k<Nf(k+1)  ≤  JN  ≤  ∑k<Nf(k)  =  sN, which is the left half of claim 1.

3.2step 2.1step 1.2givenL2L7

(JN) is nondecreasing: JN+1−JN=∫NN+1f≥0 by step 1.2 and [L2], since f≥0.

4.1step 3.1step 1.3givenL4L7

So sN+1≤f(0)+JN by step 3.1 and step 1.3; and sN≤sN+1 because f(N)≥0, so sN≤f(0)+JN, which is the right half of claim 1.

4.2step 3.1givenL5L7

If ∑kf(k) converges, then by [L5] the partial sums are bounded above, say sN≤S for every N, and step 3.1 gives JN≤sN≤S; so the set of JN is bounded above.

5.1step 4.1givenL5L7

If the set of JN is bounded above, say by a real B, then sN≤f(0)+B for every N by step 4.1, so the partial sums are bounded above and ∑kf(k) converges by [L5], the terms f(k) being nonnegative.

6.1step 3.1step 4.1step 3.2step 5.1step 4.2∎

Steps 5.1 and 4.2 are the two implications of claim 2, and step 3.2 is its first clause; claim 1 is steps 3.1 and 4.1.

Remarks

Depends on

Dependency tree · 2 levels

Sources