Alphabeta Math
TheoremStatement: Literature-sourcedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum

Statement

Let (ak) be a sequence of reals with ak≥0 for every k∈N, let sn=∑k<nak be its partial sums and let S={ sn:n∈N } be the range of (sn) (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

  1. (sn) is nondecreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and sn≥0 for every n;
  2. ∑ak converges if and only if S is bounded above (Lower bound, bounded below, bounded set), and in that case ∑k=0∞ak  =  sup⁡S, so in particular sn≤∑k=0∞ak for every n;
  3. if S is not bounded above then sn→+∞ (Divergence to +∞ and to −∞) and ∑ak diverges.

This is the theorem that makes the nonnegative theory work: for terms of one sign, convergence is a boundedness question and no candidate limit is ever needed. Every comparison test on this page is an application of it.

Facts & Assumptions

Given: A sequence (ak) of reals with ak≥0 for every k, its partial sums sn=∑k<nak, and the range S={sn:n∈N} (Series, partial sums, convergence and the sum, divergence, and the tail series, Finite sums and finite products, by recursion, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L1]

The recursion clause of the finite sum: sn+1=sn+an (Finite sums and finite products, by recursion).

[L2]

Consecutive comparisons suffice for monotonicity: (xk) is nondecreasing if and only if xk≤xk+1 for every k; and a nondecreasing sequence is bounded below by its first term (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

Monotonicity of finite sums: if ak≥0 for all k<n then ∑k<nak≥0 (Laws of finite sums and finite products).

[L4]

A monotone sequence converges if and only if it is bounded, that is if and only if there is M with ∣xk∣≤M for every k (A monotone sequence converges if and only if it is bounded, Sequences of reals: bounded, eventually, frequently, tails, subsequences).

[L6]

A nondecreasing sequence whose range is not bounded above diverges to +∞ (A nondecreasing sequence that is not bounded above diverges to +∞, Divergence to +∞ and to −∞).

Proof

technique · direct
1.1

For every n, sn+1−sn=an≥0, so sn≤sn+1 and (sn) is nondecreasing.

givenL1L2
1.2

For every n, sn=∑k<nak≥0, all terms being nonnegative.

givenL3
2.1

Claim 1 is steps 1.1 and 1.2 together.

step 1.1step 1.2
2.2

Since sn≥0 we have ∣sn∣=sn, so (sn) is bounded in the sense of [L4] if and only if S is bounded above.

step 1.2L4
3.1

By [L4] applied to the monotone sequence (sn), the series converges if and only if (sn) is bounded, hence if and only if S is bounded above.

step 1.1step 2.2L4
4.1

If S is bounded above then (sn) converges to sup⁡S, so ∑ak converges with sum sup⁡S; and since sup⁡S is an upper bound of S, sn≤sup⁡S for every n.

step 1.1step 3.1L5
4.2

If S is not bounded above then sn→+∞, and by step 3.1 the series diverges.

step 1.1step 3.1L6
5.1

The equivalence and the identification of the sum as the supremum together make claim 2, and the divergence statement is claim 3.

step 3.1step 4.1step 4.2∎

Remarks

Depends on

Used by

…and 14 more results.

Dependency tree · two levels

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Sources