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LemmaStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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If ak\sum |a_k| converges then ak\sum a_k converges

Statement

Let (ak)(a_k) be a sequence of reals. If the series ak\sum |a_k| converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then the series ak\sum a_k converges.

A series with the property that ak\sum |a_k| converges is called absolutely convergent; the lemma says that absolute convergence implies convergence.

The same statement holds for a family from a general starting index mm, being this statement applied to the shifted sequence jaj+mj \mapsto a_{j+m} (Series, partial sums, convergence and the sum, divergence, and the tail series).

The converse is false, and the standard witness is the alternating harmonic series. That witness is not available on this page: its convergence is the alternating series test, which is not proved here. Nothing below asserts a converse, and no item on this page uses one.

Facts & Assumptions

Given: A sequence (ak)(a_k) of reals such that the series ak\sum |a_k| converges, with partial sums as in Series, partial sums, convergence and the sum, divergence, and the tail series and finite sums as in Finite sums and finite products, by recursion.

[L1]

The Cauchy criterion for series: dk\sum d_k converges if and only if for every real ε>0\varepsilon > 0 there is NNN \in \mathbb{N} with k=p+1ndk<ε\big|\sum_{k=p+1}^{n} d_k\big| < \varepsilon for all n>pNn > p \ge N (A series converges iff for every ε>0\varepsilon > 0 there is NN with am+1++an<ε|a_{m+1} + \dots + a_n| < \varepsilon for all n>mNn > m \ge N, Series, partial sums, convergence and the sum, divergence, and the tail series).

[L2]

Triangle inequality for finite sums: k<dukk<duk\big|\sum_{k<d} u_k\big| \le \sum_{k<d}|u_k| (Triangle inequality for finite sums); the block k=p+1nuk\sum_{k=p+1}^{n} u_k is by definition the finite sum t<npup+1+t\sum_{t < n-p} u_{p+1+t} (Finite sums and finite products, by recursion), so applying the inequality to the shifted sequence tup+1+tt \mapsto u_{p+1+t} gives k=p+1nukk=p+1nuk\big|\sum_{k=p+1}^{n} u_k\big| \le \sum_{k=p+1}^{n} |u_k| for all naturals n>pn > p.

[L3]

Monotonicity of finite sums: if xt0x_t \ge 0 for all t<dt < d then t<dxt0\sum_{t<d} x_t \ge 0 (Laws of finite sums and finite products).

[L4]

Absolute value: u0|u| \ge 0 for every real uu, and u=u|u| = u whenever u0u \ge 0 (Basic properties of the absolute value).

[L5]

Convergence of a real sequence, and the fact that the real and rational formulations of a tolerance agree (Limits and Cauchy sequences of reals).

Proof

technique · direct
1.1

Let ε>0\varepsilon > 0 be an arbitrary real; since ak\sum |a_k| converges, the Cauchy criterion applied to the sequence (ak)(|a_k|) supplies NNN \in \mathbb{N} with k=p+1nak<ε\big|\sum_{k=p+1}^{n} |a_k|\big| < \varepsilon for all n>pNn > p \ge N.

L1L5choose
1.2

For all naturals n>pn > p the block k=p+1nak\sum_{k=p+1}^{n} |a_k| is a finite sum of nonnegative terms, hence nonnegative, hence equal to its own absolute value.

L2L3L4
2.1

So for all n>pNn > p \ge N one has k=p+1nakk=p+1nak=k=p+1nak<ε\big|\sum_{k=p+1}^{n} a_k\big| \le \sum_{k=p+1}^{n} |a_k| = \big|\sum_{k=p+1}^{n} |a_k|\big| < \varepsilon.

step 1.1step 1.2L2
3.1

As ε>0\varepsilon > 0 was arbitrary, the sequence (ak)(a_k) satisfies the Cauchy criterion, so ak\sum a_k converges.

step 2.1L1

Remarks

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 52 results over 12 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