False statementConstruction: AI-adaptedVerification: AI-generatedprecheck 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 $(a_{k})$ of reals, if $\sum a_{k}$ converges (Series, partial sums, convergence and the sum, divergence, and the tail series) then $\sum a_{k}$ converges absolutely (Absolutely convergent and conditionally convergent series, and the general starting index).

What is true is the converse, If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges: absolute convergence implies convergence. The claim above reverses it, and the reversal fails at the standard witness, the alternating harmonic series.

Let $(ε_{j})$ 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 $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ), usually written $ε_{j} = (- 1)^{j}$ , and put

$a_{j} : = \frac{ε_{j}}{ι (j + 1)} (j \in N),$

with $ι (j + 1)$ the canonical natural (Canonical naturals are positive and strictly increasing). Then $\sum a_{j}$ converges while $\sum ∣ a_{j} ∣$ 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 $(ε_{j})$ , the sequence $b_{j} : = 1 / ι (j + 1)$ , and $a_{j} : = ε_{j} b_{j}$ .

[A1]

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

[L1]

$∣ ε_{j} ∣ = 1$ for every $j$ (The even and odd index maps and the alternating sequence: strictly increasing $e, o$ with $N$ their disjoint union, and the unique $(s_{k})$ with $s_{0} = 1$ , $s_{σ (k)} = - s_{k}$ , which satisfies $∣ s_{k} ∣ = 1$ , $s \circ e \equiv 1$ and $s \circ o \equiv - 1$ ).

[L2]

The canonical naturals $ι (n)$ are positive for $n \geq 1$ and strictly increasing in $n$ (Canonical naturals are positive and strictly increasing).

[L3]

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

[L4]

For every real $ε > 0$ there is a natural $n \geq 1$ with $1 / ι (n) < ε$ (For every $ε > 0$ in a complete ordered field there is a natural $n \geq 1$ with $1 / n < ε$ ).

[L5]

The alternating series test: if $(b_{j})$ is nonincreasing (Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences) with $b_{j} \to 0$ , then $\sum ε_{j} b_{j}$ converges (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 $b_{n}$ , Limits and Cauchy sequences of reals).

[L6]

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

[L7]

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

[L8]

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

[L9]

Absolute convergence means convergence of $\sum ∣ a_{j} ∣$ ; conditional convergence means convergence of $\sum a_{j}$ without it (Absolutely convergent and conditionally convergent series, and the general starting index).

[L10]

Absolute convergence implies convergence (If $\sum ∣ a_{k} ∣$ converges then $\sum a_{k}$ converges).

Refutation

technique · direct

1.1givenL2

Each $b_{j} = 1 / ι (j + 1)$ is a positive real, $ι (j + 1)$ being a positive canonical natural.

1.2L2L3

The sequence $(b_{j})$ is nonincreasing: $ι (j + 1) < ι (j + 2)$ , so $1 / ι (j + 2) < 1 / ι (j + 1)$ .

2.1step 1.1L2L3L4

The sequence $(b_{j})$ converges to $0$ : given a rational $ε > 0$ , fix a natural $n \geq 1$ with $1 / ι (n) < ε$ ; then for every $j \geq n$ one has $ι (j + 1) \geq ι (n) > 0$ , hence $∣ b_{j} ∣ = b_{j} \leq 1 / ι (n) < ε$ .

2.2step 1.1L1L8

For every $j$ , $∣ a_{j} ∣ = ∣ ε_{j} ∣ ∣ b_{j} ∣ = b_{j} = 1 / ι (j + 1)$ .

3.1step 1.2step 2.1L5

By the alternating series test, $\sum a_{j} = \sum ε_{j} b_{j}$ converges.

3.2step 2.2L6L7

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

4.1step 3.2L6

The $p$ -series at $p = 1$ diverges, since $1 > 1$ is false; so $\sum ∣ a_{j} ∣$ diverges.

5.1step 3.1step 4.1A1L9

Thus $\sum a_{j}$ converges while $\sum ∣ a_{j} ∣$ does not, so $\sum a_{j}$ converges conditionally and not absolutely, and the claim [A1] fails for this series.

6.1step 5.1A1L10∎

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

Remarks

The witness is not exotic. It is the first series a reader meets whose convergence depends on cancellation, and the failure is as large as it can be: the series of absolute values does not merely converge to a different number, it diverges to $+ \infty$ .
Everything on this page turns on this example. Because the class of conditionally convergent series is nonempty, The Riemann series theorem: a conditionally convergent real series has, for every $c \in R$ , a rearrangement with sum $c$ , and rearrangements diverging to $+ \infty$ , to $- \infty$ , and oscillating with any prescribed $lim inf \leq lim sup$ in $\overline{R}$ has content, and For a series of real numbers, unconditional convergence and absolute convergence are the same property separates two genuinely different properties rather than restating one. The same series, with its rearrangements, is developed on the companion page.
The value of the sum is not asserted here. The series is proved convergent and nothing more; the classical evaluation needs the logarithm, which is later in the reading order (Selected sums and products on this page that are proved to exist without being evaluated, and what their evaluation waits for).

FALSE: every convergent series converges absolutely

Statement

Facts & Assumptions

Refutation

Remarks

Depends on

Dependency tree · 2 levels

Sources