TheoremStatement: Literature-sourcedProof: AI-adaptedSession-authored (Fable 5 assisted)precheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

How statement and proof provenance work

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.
AI-generated — a genuinely novel statement formulated by AI, with no source for the claim itself.

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

The alternating series test: if $(b_k)$ is nonincreasing with $b_k \to 0$ then $\sum_{k} (-1)^{k} b_k$ converges, the sum lies between any two consecutive partial sums, and the error after $n$ terms is at most $b_n$

Statement

Let $(\varepsilon_k)$ be the alternating sequence of The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ , that is the unique sequence of reals with $\varepsilon_0 = 1$ and $\varepsilon_{k+1} = -\varepsilon_k$ , which is what is usually written $\varepsilon_k = (-1)^k$ ; let $e$ and $o$ be its even and odd index maps, so that $\varepsilon_{e_j} = 1$ , $\varepsilon_{o_j} = -1$ , and every natural number is $e_j$ for exactly one $j$ or $o_j$ for exactly one $j$ .

Let $(b_k)$ be a sequence of reals that is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) and converges to $0$ (Limits and Cauchy sequences of reals); then $b_k \ge 0$ for every $k$ . Write $t_n := \sum_{k<n} \varepsilon_k b_k$ for the partial sums (Series, partial sums, convergence and the sum, divergence, and the tail series). Then:

the series $\sum \varepsilon_k b_k$ converges; write $L$ for its sum;
$t_{e_j} \le L \le t_{o_j}$ for every $j \in \mathbb{N}$ , and for every $n \in \mathbb{N}$ the sum $L$ lies between the two consecutive partial sums $t_n$ and $t_{n+1}$ ;
$|L - t_n| \le b_n$ for every $n \in \mathbb{N}$ .

Claim 3 is the error bound: the partial sum $t_n$ , which uses the $n$ terms $\varepsilon_0 b_0, \dots, \varepsilon_{n-1}b_{n-1}$ , differs from the sum by at most the first term omitted.

Only claim 1 is a corollary of Dirichlet's test: if the partial sums of $\sum a_k$ are bounded and $(b_k)$ is nonincreasing with $b_k \to 0$ , then $\sum a_k b_k$ converges. Claims 2 and 3 are not: they come from the interlacing of the even-index and odd-index partial sums, and that argument is carried out below rather than smuggled into the Dirichlet estimate, which produces no bracketing at all.

Facts & Assumptions

Given: A nonincreasing sequence $(b_k)$ of reals with $b_k \to 0$ , the alternating sequence $(\varepsilon_k)$ with its index maps $e$ and $o$ , and the partial sums $t_n = \sum_{k<n} \varepsilon_k b_k$ .

[L1]

The alternating sequence and its index maps: $\varepsilon_0 = 1$ , $\varepsilon_{k+1} = -\varepsilon_k$ , $|\varepsilon_k| = 1$ ; $e_0 = 0$ and $e_{j+1} = e_j + 2$ ; $o_0 = 1$ and $o_{j+1} = o_j + 2$ ; both $e$ and $o$ are strictly increasing; $\mathbb{N}$ is the disjoint union of their ranges; $\varepsilon_{e_j} = 1$ and $\varepsilon_{o_j} = -1$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ ).

[L2]

Nonincreasing means $b_j \ge b_k$ whenever $j \le k$ (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L3]

Limits preserve non-strict inequalities holding eventually (Limits preserve non-strict inequalities, Limits and Cauchy sequences of reals).

[L4]

Dirichlet's test: if the partial sums of $\sum x_k$ are bounded and $(y_k)$ is nonincreasing with $y_k \to 0$ , then $\sum x_k y_k$ converges (Dirichlet's test: if the partial sums of $\sum a_k$ are bounded and $(b_k)$ is nonincreasing with $b_k \to 0$ , then $\sum a_k b_k$ converges).

[L5]

A subsequence of a convergent sequence converges to the same limit (Subsequences inherit the limit).

[L6]

Partial sums satisfy $t_0 = 0$ and $t_{n+1} = t_n + \varepsilon_n b_n$ (Series, partial sums, convergence and the sum, divergence, and the tail series).

[L7]

The principle of induction on $\mathbb{N}$ (The principle of mathematical induction).

[L8]

Absolute value: $|xy| = |x|\,|y|$ and $|x| \ge 0$ (Basic properties of the absolute value).

Proof

technique · direct

1.1

For each fixed $k$ the inequality $b_m \le b_k$ holds for all $m \ge k$ , and $(b_m)_m$ converges to $0$ while the constant sequence with value $b_k$ converges to $b_k$ ; hence $b_k \ge 0$ .

givenL2L3

1.2

Writing $A_n = \sum_{k<n}\varepsilon_k$ , an induction gives that for every $n$ either $A_n = 0$ and $\varepsilon_n = 1$ , or $A_n = 1$ and $\varepsilon_n = -1$ : at $n = 0$ we have $A_0 = 0$ and $\varepsilon_0 = 1$ ; and if $A_n = 0$ and $\varepsilon_n = 1$ then $A_{n+1} = 1$ and $\varepsilon_{n+1} = -1$ , while if $A_n = 1$ and $\varepsilon_n = -1$ then $A_{n+1} = 0$ and $\varepsilon_{n+1} = 1$ . In particular $|A_n| \le 1$ for every $n$ .

L1L6L7

1.3

For every $j$ one has $o_j = e_j + 1$ and $e_{j+1} = o_j + 1$ , by induction: $o_0 = 1 = e_0 + 1$ ; and if $o_j = e_j + 1$ then $e_{j+1} = e_j + 2 = o_j + 1$ and $o_{j+1} = o_j + 2 = e_{j+1} + 1$ .

L1L7

1.4

By [L6], $t_{n+1} - t_n = \varepsilon_n b_n$ for every $n$ ; hence $t_{e_j + 1} = t_{e_j} + b_{e_j}$ and $t_{o_j + 1} = t_{o_j} - b_{o_j}$ .

L1L6

2.1

The partial sums of $\sum \varepsilon_k$ are bounded by step 1.2 and $(b_k)$ is nonincreasing with limit $0$ , so $\sum \varepsilon_k b_k$ converges by Dirichlet's test; write $L$ for its sum, so that $t_n \to L$ .

step 1.2givenL4

2.2

Using step 1.3, $t_{o_j} = t_{e_j + 1} = t_{e_j} + b_{e_j}$ and $t_{e_{j+1}} = t_{o_j + 1} = t_{o_j} - b_{o_j}$ , so $t_{e_{j+1}} = t_{e_j} + b_{e_j} - b_{o_j}$ and $t_{o_{j+1}} = t_{e_{j+1}} + b_{e_{j+1}} = t_{o_j} - b_{o_j} + b_{e_{j+1}}$ .

step 1.3step 1.4

3.1

Since $e_j < o_j < e_{j+1}$ and $(b_k)$ is nonincreasing, $b_{e_j} - b_{o_j} \ge 0$ and $b_{e_{j+1}} - b_{o_j} \le 0$ ; so by step 2.2 the sequence $(t_{e_j})_j$ is nondecreasing and the sequence $(t_{o_j})_j$ is nonincreasing.

step 1.3step 2.2L2

3.2

The maps $e$ and $o$ are strictly increasing, so $(t_{e_j})_j$ and $(t_{o_j})_j$ are subsequences of $(t_n)$ and both converge to $L$ .

step 2.1L1L5

4.1

Fix $j$ . For every $m \ge j$ one has $t_{e_j} \le t_{e_m}$ , and $(t_{e_m})_m$ converges to $L$ , so $t_{e_j} \le L$ ; symmetrically $t_{o_j} \ge L$ . This is the first half of claim 2.

step 3.1step 3.2L3

5.1

Let $n \in \mathbb{N}$ . If $n = e_j$ then $t_n = t_{e_j} \le L$ and $t_{n+1} = t_{e_j+1} = t_{o_j} \ge L$ ; if $n = o_j$ then $t_n = t_{o_j} \ge L$ and $t_{n+1} = t_{o_j+1} = t_{e_{j+1}} \le L$ . Since every $n$ is of exactly one of these two forms, $L$ always lies between $t_n$ and $t_{n+1}$ , which is the second half of claim 2.

step 1.3step 4.1L1

6.1

Consequently $|L - t_n| \le |t_{n+1} - t_n| = |\varepsilon_n b_n| = |\varepsilon_n|\,b_n = b_n$ for every $n$ , using $b_n \ge 0$ and $|\varepsilon_n| = 1$ ; this is claim 3.

step 5.1step 1.4step 1.1L1L8∎

Remarks

The two hypotheses are not interchangeable with " $b_k \to 0$ " alone. A null sequence that is not monotone can make $\sum \varepsilon_k b_k$ diverge, and the bracketing of step 3.1 is exactly where monotonicity enters; the error bound is false without it. The test as stated is the classical Leibniz criterion.
Why the index maps rather than " $2j$ " and " $2j+1$ ". The even and odd index maps come from The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $\mathbb{N}$ their disjoint union, and the unique $(s_k)$ with $s_0 = 1$ , $s_{\sigma(k)} = -s_k$ , which satisfies $|s_k| = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv -1$ together with the parity object itself, and step 1.3 is the only arithmetic needed about them. Rebuilding $(-1)^k$ by a fresh recursion inside this proof, and then proving afresh that the even indices and the odd indices partition $\mathbb{N}$ , is precisely what that lemma exists to prevent.
What the test does not give. It produces the sum $L$ as a limit and bounds the error, and it identifies $L$ with no closed expression. For the alternating harmonic series the value is not available at this point in the reading order; see Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for.

Depends on

Used by

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 84 results over 18 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

Alternating series test (Wikipedia) (standard reference, not scraped)
W. Rudin, Principles of Mathematical Analysis, 3rd ed., Ch. 3 (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)

The alternating series test: if (bk)(b_k)(bk​) is nonincreasing with bk→0b_k \to 0bk​→0 then ∑k(−1)kbk\sum_{k} (-1)^{k} b_k∑k​(−1)kbk​ converges, the sum lies between any two consecutive partial sums, and the error after nnn terms is at most bnb_nbn​