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∑j≥0(−1)j/(j+1) converges conditionally, with sum strictly between 1/2 and 1

Example

Let (εj) be the alternating sequence (The even and odd index maps and the alternating sequence: strictly increasing e,o with N their disjoint union, and the unique (sk) with s0=1, sσ(k)=−sk, which satisfies ∣sk∣=1, s∘e≡1 and s∘o≡−1), written εj=(−1)j, and put bj:=1/ι(j+1), with ι(j+1) the canonical natural, positive for every j (Canonical naturals are positive and strictly increasing). The alternating harmonic series is

∑j≥0(−1)jj+1  =  ∑jεjbj.

It converges conditionally (Absolutely convergent and conditionally convergent series, and the general starting index): it converges, by the alternating series test, while its series of absolute values is the harmonic series ∑k≥11/k, which diverges (For rational p>0, ∑1/kp converges iff p>1). Writing S for its sum,

12  <  712  ≤  S  ≤  56  <  1.

The value of S is not asserted. The classical evaluation is a logarithm and is not available at this point in the reading order; what is proved here is that S exists and where it lies. See Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for.

This is the series that gives the whole page its content: it is the standard witness for FALSE: every convergent series converges absolutely and, through The Riemann series theorem: a conditionally convergent real series has, for every c∈R, a rearrangement with sum c, and rearrangements diverging to +∞, to −∞, and oscillating with any prescribed lim inf⁡≤lim sup⁡ in R‾, the source of every rearrangement example below.

Facts & Assumptions

Given: The alternating sequence (εj) with index maps e and o, the sequence bj=1/ι(j+1), and the partial sums tn=∑j<nεjbj (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

The alternating sequence: ε0=1, εj+1=−εj, ∣εj∣=1; e0=0, ej+1=ej+2, o0=1, oj+1=oj+2; εej=1 and εoj=−1 (The even and odd index maps and the alternating sequence: strictly increasing e,o with N their disjoint union, and the unique (sk) with s0=1, sσ(k)=−sk, which satisfies ∣sk∣=1, s∘e≡1 and s∘o≡−1).

[L2]

The canonical naturals are positive for n≥1 and strictly increasing; reciprocation reverses the order on the positives; and for every real ε>0 there is n≥1 with 1/ι(n)<ε (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, For every ε>0 in a complete ordered field there is a natural n≥1 with 1/n<ε).

[L5]

Absolute value: ∣xy∣=∣x∣ ∣y∣ (Basic properties of the absolute value).

[L7]

Verification

technique · direct
1.1

Every bj=1/ι(j+1) is positive, and (bj) is nonincreasing, since 0<ι(j+1)<ι(j+2).

givenL2
1.2

By [L1], e1=2, e2=4, o1=3; and by [L6] together with εej=1, εoj=−1, the first partial sums are t1=b0=1, t2=1−1/2=1/2, t3=1/2+1/3=5/6 and t4=5/6−1/4=7/12.

L1L6algebra
2.1

(bj) converges to 0: given a rational ε>0, take n≥1 with 1/ι(n)<ε; for j≥n one has ι(j+1)≥ι(n)>0, so bj≤1/ι(n)<ε.

step 1.1L2
2.2

For every j, ∣εjbj∣=∣εj∣ bj=1/ι(j+1), and ∑j1/ι(j+1) is the p-series ∑k≥11/k at p=1, which diverges.

step 1.1L1L4L5
3.1

By the alternating series test the series converges; write S for its sum, and tej≤S≤toj holds for every j.

step 1.1step 2.1L3
4.1

Taking j=2 in the lower bound and j=1 in the upper bound of step 3.1 gives 7/12=t4=te2≤S≤to1=t3=5/6.

step 3.1step 1.2
5.1

Since 1/2<7/12 and 5/6<1, the sum satisfies 1/2<S<1.

step 4.1algebra
6.1

So the series converges while its series of absolute values diverges: it converges conditionally, with sum strictly between 1/2 and 1.

step 3.1step 5.1step 2.2L7∎

Remarks

  • The bracketing is exactly the error bound of the test, used twice. Any pair of an even-index and an odd-index partial sum brackets S, and the further out the pair is taken the tighter the bracket becomes; t4 and t3 are simply the first pair whose values separate S strictly from 1/2 and from 1. Taking t2=1/2 and t1=1 would give only the non-strict bounds.

  • Conditional convergence is a statement about cancellation. The terms have absolute value 1/(j+1) and their sum without signs is infinite; the series converges only because consecutive terms nearly cancel. Everything that follows on this page, that the terms may be reordered to sum to anything at all, is a consequence of exactly that.

  • What the bracket does not say. It gives no rate and no closed form. Better numerical bounds come from later pairs tej,toj and cost only arithmetic; the closed form costs the logarithm.

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Sources