CounterexampleConstruction: AI-adaptedVerification: AI-generatedprecheck passjudge pass (z-ai/glm-5.2)audited 2026-07-27

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A nonnegative non-monotone sequence for which $\sum a_{k}$ and $\sum 2^{k} a_{2^{k}}$ behave differently

Statement refuted

Refuted claim: for every family $(a_{k})_{k \geq 1}$ with $a_{k} \geq 0$ , $\sum_{k \geq 1} a_{k}$ converges if and only if $\sum_{j \geq 0} 2^{j} a_{2^{j}}$ converges.

This is For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges with its monotonicity hypothesis deleted. Let $P : = {2^{j} : j \in N}$ be the set of powers of $2$ and define, for naturals $k \geq 1$ ,

$a_{k} : = {\begin{cases} 0 & if k \in P, \\ 1 & if k \notin P . \end{cases}$

Every term is nonnegative, and the family is not monotone in either direction: $a_{1} = 0 < 1 = a_{3}$ and $a_{3} = 1 > 0 = a_{4}$ , since $1 = 2^{0}$ and $4 = 2^{2}$ belong to $P$ while $3$ does not.

The condensed series is $\sum_{j \geq 0} 2^{j} a_{2^{j}} = \sum_{j \geq 0} 0$ , which converges with sum $0$ . The original series $\sum_{k \geq 1} a_{k}$ diverges, because $a_{k} = 1$ at arbitrarily large indices, so its terms do not tend to $0$ (If a series converges then its terms tend to $0$ ).

Facts & Assumptions

Given: $P = {2^{j} : j \in N}$ and the family $a_{k}$ defined above for naturals $k \geq 1$ (Series, partial sums, convergence and the sum, divergence, and the tail series, Integer powers $a^{m}$ ).

[L1]

Powers of $2$ : $2^{j} \geq 1$ , $2^{j + 1} = 2 \cdot 2^{j}$ , and $j \mapsto 2^{j}$ is strictly increasing, since $2 > 1$ (Integer powers $a^{m}$ , Monotonicity of $x \mapsto x^{n}$ and of $n \mapsto a^{n}$ , Canonical naturals are positive and strictly increasing).

[L2]

The naturals are discrete: no natural lies strictly between $n$ and $n + 1$ (Discreteness: $σ (n)$ is the immediate successor).

[L3]

The principle of induction (The principle of mathematical induction).

[L4]

A finite sum of zeros is zero, and a constant sequence converges to its value (Laws of finite sums and finite products, Finite sums and finite products, by recursion, Limits and Cauchy sequences of reals).

[L5]

A series whose terms do not converge to $0$ diverges (If a series converges then its terms tend to $0$ , Limits and Cauchy sequences of reals).

[L6]

Condensation requires the family to be nonnegative and nonincreasing (For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges, Nondecreasing, increasing, nonincreasing, decreasing, monotone, and eventually monotone sequences).

[L7]

The refuted claim: nonnegativity alone suffices for the condensation equivalence.

Counterexample

technique · direct

1.1

Every $a_{k}$ is $0$ or $1$ , hence nonnegative, so the family satisfies the hypothesis of the claim.

givenL7

1.2

The family is not monotone: $1 = 2^{0} \in P$ and $4 = 2^{2} \in P$ give $a_{1} = a_{4} = 0$ , while $3 \notin P$ gives $a_{3} = 1$ ; so $a_{1} < a_{3}$ rules out nonincreasing and $a_{3} > a_{4}$ rules out nondecreasing. That $3 \notin P$ holds because $2^{1} = 2 < 3 < 4 = 2^{2}$ and $j \mapsto 2^{j}$ is strictly increasing, so a power of $2$ equal to $3$ would force a natural strictly between $1$ and $2$ .

givenL1L2

1.3

Every condensed term is $2^{j} a_{2^{j}} = 2^{j} \cdot 0 = 0$ , since $2^{j} \in P$ for every $j$ .

givenL1

1.4

An induction gives $2^{n} > ι (n)$ for every $n \in N$ : at $n = 0$ this reads $1 > 0$ ; and if $2^{n} > ι (n)$ then $2^{n + 1} = 2^{n} + 2^{n} \geq 2^{n} + 1 > ι (n) + 1 = ι (n + 1)$ .

L1L3

1.5

For every $n \geq 1$ the natural $2^{n} + 1$ is not in $P$ : it satisfies $2^{n} < 2^{n} + 1 < 2^{n + 1}$ , the second inequality because $2^{n + 1} = 2^{n} + 2^{n} \geq 2^{n} + 2$ ; so a power of $2$ equal to it would force a natural strictly between $n$ and $n + 1$ .

L1L2

2.1

So the condensed series has all partial sums equal to $0$ and converges, with sum $0$ .

step 1.3L4

2.2

Hence for every $n \geq 1$ the index $k : = 2^{n} + 1$ satisfies $k > ι (n) \geq n$ and $a_{k} = 1$ , so $a_{k} = 1$ at indices exceeding any prescribed bound.

step 1.4step 1.5given

3.1

Therefore the terms of $\sum_{k \geq 1} a_{k}$ do not converge to $0$ : with the rational tolerance $1 / 2$ no index $K$ satisfies $∣ a_{k} ∣ < 1 / 2$ for all $k \geq K$ . So that series diverges.

step 2.2L5

4.1

The condensed series converges while the original diverges, so the claimed equivalence fails and the claim is false; the genuine condensation theorem is untouched, since its monotonicity hypothesis is violated here.

step 2.1step 3.1step 1.2L6L7∎

Remarks

The witness knocks out exactly one estimate. Condensation squeezes the block $a_{2^{n}}, \dots, a_{2^{n + 1} - 1}$ between $2^{n}$ copies of its last term and $2^{n}$ copies of its first, and both bounds are consequences of monotonicity. Here the first term of each block is $0$ and the rest are $1$ , so the upper bound $2^{n} a_{2^{n}} = 0$ is wildly wrong, and it is the upper bound that the convergence direction of the theorem uses.
The failure is one-directional here, and the other direction can fail too. This witness has a convergent condensed series and a divergent original. The complementary family $a_{k} : = 1 / k$ for $k \in P$ and $a_{k} : = 0$ otherwise reverses the roles, its original series being a geometric one and its condensed series having every term equal to $1$ ; that variant is not verified here, and only the direction exhibited above is claimed.

Depends on

Used by

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Sources

Cauchy condensation test (Wikipedia) (standard reference, not scraped)
Stephen Semmes, Elements of Analysis (standard reference, not scraped)

A nonnegative non-monotone sequence for which ∑ak∑ak​ and ∑2ka2k∑2ka2k​ behave differently