Alphabeta Math
False statementConstruction: AI-adaptedVerification: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27
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FALSE: every convergent series converges absolutely

Statement

False claim: for every sequence (ak)(a_k) of reals, if ak\sum a_k converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then ak\sum a_k converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index).

What is true is the converse, If ak\sum |a_k| converges then ak\sum a_k converges: absolute convergence implies convergence. The claim above reverses it, and the reversal fails at the standard witness, the alternating harmonic series.

Let (εj)(\varepsilon_j) be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing e,oe, o with N\mathbb{N} their disjoint union, and the unique (sk)(s_k) with s0=1s_0 = 1, sσ(k)=sks_{\sigma(k)} = -s_k, which satisfies sk=1|s_k| = 1, se1s \circ e \equiv 1 and so1s \circ o \equiv -1), usually written εj=(1)j\varepsilon_j = (-1)^j, and put

aj  :=  εjι(j+1)(jN),a_j \;:=\; \frac{\varepsilon_j}{\iota(j+1)} \qquad (j \in \mathbb{N}),

with ι(j+1)\iota(j+1) the canonical natural (Canonical naturals are positive and strictly increasing). Then aj\sum a_j converges while aj\sum |a_j| is the harmonic series, which diverges. So the two notions really are different, and "conditionally convergent" is not an empty class.

Facts & Assumptions

Given: The alternating sequence (εj)(\varepsilon_j), the sequence bj:=1/ι(j+1)b_j := 1/\iota(j+1), and aj:=εjbja_j := \varepsilon_j b_j.

[A1]

The refuted claim: every convergent series of reals converges absolutely.

[L2]

The canonical naturals ι(n)\iota(n) are positive for n1n \ge 1 and strictly increasing in nn (Canonical naturals are positive and strictly increasing).

[L3]

If 0<u<v0 < u < v then 0<1/v<1/u0 < 1/v < 1/u (Inverses of positives are positive, and reciprocation reverses order).

[L4]

For every real ε>0\varepsilon > 0 there is a natural n1n \ge 1 with 1/ι(n)<ε1/\iota(n) < \varepsilon (For every ε>0\varepsilon > 0 in a complete ordered field there is a natural n1n \ge 1 with 1/n<ε1/n < \varepsilon).

[L6]

k11/kp\sum_{k \ge 1} 1/k^{p} converges if and only if p>1p > 1, where kp=ι(k)pk^{p} = \iota(k)^{p}; at p=1p = 1 the rational power is the element itself, ι(k)1=ι(k)\iota(k)^{1} = \iota(k) (For rational p>0p > 0, 1/kp\sum 1/k^p converges iff p>1p > 1, Rational powers ara^r of a positive base, Existence and uniqueness of nn-th roots: a unique a1/n0a^{1/n} \ge 0 with (a1/n)n=a(a^{1/n})^n = a, Integer powers ama^m).

[L7]

The series k1xk\sum_{k \ge 1} x_k is by definition the series of the sequence jxj+1j \mapsto x_{j+1} (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L8]

Absolute value: xy=xy|xy| = |x|\,|y| (Basic properties of the absolute value).

[L9]

Absolute convergence means convergence of aj\sum |a_j|; conditional convergence means convergence of aj\sum a_j without it (Absolutely convergent and conditionally convergent series, and the general starting index).

Refutation

technique · direct
1.1

Each bj=1/ι(j+1)b_j = 1/\iota(j+1) is a positive real, ι(j+1)\iota(j+1) being a positive canonical natural.

givenL2
1.2

The sequence (bj)(b_j) is nonincreasing: ι(j+1)<ι(j+2)\iota(j+1) < \iota(j+2), so 1/ι(j+2)<1/ι(j+1)1/\iota(j+2) < 1/\iota(j+1).

L2L3
2.1

The sequence (bj)(b_j) converges to 00: given a rational ε>0\varepsilon > 0, fix a natural n1n \ge 1 with 1/ι(n)<ε1/\iota(n) < \varepsilon; then for every jnj \ge n one has ι(j+1)ι(n)>0\iota(j+1) \ge \iota(n) > 0, hence bj=bj1/ι(n)<ε|b_j| = b_j \le 1/\iota(n) < \varepsilon.

step 1.1L2L3L4
2.2

For every jj, aj=εjbj=bj=1/ι(j+1)|a_j| = |\varepsilon_j|\,|b_j| = b_j = 1/\iota(j+1).

step 1.1L1L8
3.1

By the alternating series test, aj=εjbj\sum a_j = \sum \varepsilon_j b_j converges.

step 1.2step 2.1L5
3.2

The series j1/ι(j+1)\sum_j 1/\iota(j+1) is, by the definition of a series from a general starting index, exactly the series k11/k\sum_{k \ge 1} 1/k, that is the pp-series at p=1p = 1.

step 2.2L6L7
4.1

The pp-series at p=1p = 1 diverges, since 1>11 > 1 is false; so aj\sum |a_j| diverges.

step 3.2L6
5.1

Thus aj\sum a_j converges while aj\sum |a_j| does not, so aj\sum a_j converges conditionally and not absolutely, and the claim [A1] fails for this series.

step 3.1step 4.1A1L9
6.1

The claim is therefore false. What survives of it is only the converse implication, that an absolutely convergent series converges.

step 5.1A1L10

Remarks

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