LemmaStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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

Positive and negative parts: $a_k = a_k^{+} - a_k^{-}$ and $|a_k| = a_k^{+} + a_k^{-}$ ; a series converges absolutely iff both $\sum a_k^{+}$ and $\sum a_k^{-}$ converge, and for a conditionally convergent series both diverge to $+\infty$

Statement

Let $(a_k)$ be a sequence of reals (Series, partial sums, convergence and the sum, divergence, and the tail series) and define its positive part and negative part by

$a_k^{+} \;:=\; \frac{|a_k| + a_k}{2}, \qquad a_k^{-} \;:=\; \frac{|a_k| - a_k}{2} \qquad (k \in \mathbb{N}),$

with $|x|$ the absolute value (Absolute value in an ordered field). Then:

$a_k^{+} = \max\{a_k, 0\}$ and $a_k^{-} = \max\{-a_k, 0\}$ (Maximum and minimum of a set); in particular $a_k^{+} \ge 0$ and $a_k^{-} \ge 0$ , and $a_k = a_k^{+} - a_k^{-}, \qquad |a_k| = a_k^{+} + a_k^{-} .$
$\sum a_k$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index) if and only if both $\sum a_k^{+}$ and $\sum a_k^{-}$ converge.
If $\sum a_k$ converges conditionally, then neither $\sum a_k^{+}$ nor $\sum a_k^{-}$ converges, and the partial sums of each diverge to $+\infty$ (Divergence to $+\infty$ and to $-\infty$ ).

Claim 3 is the engine of the rearrangement theory: a conditionally convergent series carries an unlimited supply of positive terms and an unlimited supply of negative ones, and its convergence is nothing but a cancellation between them.

Facts & Assumptions

Given: A sequence $(a_k)$ of reals, its positive and negative parts $a_k^{+}$ and $a_k^{-}$ as displayed above, and the partial sums of the associated series (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L1]

Absolute value: $|x| \ge 0$ , $-|x| \le x \le |x|$ , and $|x| = x$ when $x \ge 0$ while $|x| = -x$ when $x < 0$ (Absolute value in an ordered field, Basic properties of the absolute value).

[L2]

A maximum of a subset of $\mathbb{R}$ is its greatest element, and there is at most one (Maximum and minimum of a set).

[L3]

Linearity of series: if $\sum x_k$ and $\sum y_k$ converge then so does $\sum (x_k + y_k)$ , and $\sum c\,x_k$ converges for every real $c$ (Convergent series add and scale termwise).

[L4]

Direct comparison: if $0 \le x_k \le y_k$ from some index on and $\sum y_k$ converges, then $\sum x_k$ converges (If $0 \le a_k \le b_k$ eventually, convergence of $\sum b_k$ gives convergence of $\sum a_k$ , and divergence of $\sum a_k$ gives divergence of $\sum b_k$ ).

[L5]

For a series of nonnegative terms, convergence is equivalent to the range of the partial sums being bounded above; and if that range is not bounded above then the partial sums diverge to $+\infty$ (A series of nonnegative terms converges iff its partial sums are bounded, and then the sum is their supremum, Divergence to $+\infty$ and to $-\infty$ ).

[L6]

$\sum a_k$ converges absolutely means $\sum |a_k|$ converges, and converges conditionally means it converges while $\sum |a_k|$ does not (Absolutely convergent and conditionally convergent series, and the general starting index, Limits and Cauchy sequences of reals).

Proof

technique · direct

1.1

For every $k$ , $|a_k| + a_k \ge 0$ and $|a_k| - a_k \ge 0$ , since $-|a_k| \le a_k \le |a_k|$ ; dividing by the positive real $2$ gives $a_k^{+} \ge 0$ and $a_k^{-} \ge 0$ .

L1algebra

1.2

For every $k$ , $a_k^{+} - a_k^{-} = \bigl((|a_k| + a_k) - (|a_k| - a_k)\bigr)/2 = a_k$ and $a_k^{+} + a_k^{-} = \bigl((|a_k| + a_k) + (|a_k| - a_k)\bigr)/2 = |a_k|$ .

algebra

1.3

Assume now that $\sum a_k$ converges conditionally, so $\sum a_k$ converges and $\sum |a_k|$ diverges.

2.1

If $a_k \ge 0$ then $|a_k| = a_k$ , so $a_k^{+} = a_k$ and $a_k^{-} = 0$ ; if $a_k < 0$ then $|a_k| = -a_k$ , so $a_k^{+} = 0$ and $a_k^{-} = -a_k$ . In both situations $a_k^{+}$ is the greater of $a_k$ and $0$ and $a_k^{-}$ is the greater of $-a_k$ and $0$ , which is claim 1 together with step 1.1 and step 1.2.

L1L2step 1.1step 1.2algebra

2.2

From step 1.1 and step 1.2, $0 \le a_k^{+} \le a_k^{+} + a_k^{-} = |a_k|$ and $0 \le a_k^{-} \le |a_k|$ for every $k$ .

step 1.1step 1.2algebra

2.3

If both $\sum a_k^{+}$ and $\sum a_k^{-}$ converge, then $\sum |a_k| = \sum (a_k^{+} + a_k^{-})$ converges.

step 1.2L3

3.1

If $\sum |a_k|$ converges then, by comparison with $\sum |a_k|$ using step 2.2, both $\sum a_k^{+}$ and $\sum a_k^{-}$ converge.

step 2.2L4

3.2

If $\sum a_k^{+}$ converged, then $\sum a_k^{-} = \sum \bigl(a_k^{+} + (-1)a_k\bigr)$ would converge by linearity, whence $\sum |a_k|$ would converge by step 2.3; since $\sum |a_k|$ diverges, $\sum a_k^{+}$ diverges.

step 1.3step 1.2step 2.3L3

3.3

If $\sum a_k^{-}$ converged, then $\sum a_k^{+} = \sum (a_k^{-} + a_k)$ would converge by linearity, whence again $\sum |a_k|$ would converge; since $\sum |a_k|$ diverges, $\sum a_k^{-}$ diverges.

step 1.3step 1.2step 2.3L3

4.1

Claim 2 is the conjunction of step 2.3 and step 3.1, read through the definition of absolute convergence.

step 2.3step 3.1L6

5.1

Both $\sum a_k^{+}$ and $\sum a_k^{-}$ are series of nonnegative terms by step 1.1, so each diverges only if the range of its partial sums fails to be bounded above, and then those partial sums diverge to $+\infty$ ; this is claim 3.

step 1.1step 3.2step 3.3L5∎

Remarks

The two parts are determined by the terms, with no choice anywhere. The displayed formulas define $a^{+}$ and $a^{-}$ outright, and step 2.1 identifies them with the two maxima; nothing in the proof selects one of several candidates.
Claim 3 is sharp in both directions. Absolute convergence makes both part series converge, and then $\sum_{k=0}^{\infty} a_k$ is the difference of their sums. Conditional convergence makes both part series diverge to $+\infty$ , and the difference of their partial sums is what converges. There is no third possibility for a convergent series, because claim 2 covers the case where one of them converges: if exactly one converged, $\sum a_k = \sum(a_k^{+} - a_k^{-})$ could not converge, since the sum of a convergent and a divergent series diverges.
Why $\max$ is mentioned at all. The formulas with $|a_k|$ are what the algebra uses, while $\max\{a_k, 0\}$ is what the name "positive part" means and what makes claims about signs immediate. Step 2.1 records that they agree, so either may be used later without further comment.

Depends on

Used by

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Direct dependencies and their dependencies through the next three levels: 60 results over 13 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

Absolute convergence (Wikipedia) (standard reference, not scraped)
Positive and negative parts (Wikipedia) (standard reference, not scraped)
John K. Hunter, An Introduction to Real Analysis (standard reference, not scraped)
N. Donaldson, Math 140A: Series (standard reference, not scraped)