LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-6-luna)

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The zeta sums at 2 and 3 are finite

Statement

For $p \in {2, 3}$ , let $ζ_{p} : = \sum_{k = 1}^{\infty} k^{- p}$ , where each positive integer in the summand is identified with its canonical real image. Both sums converge and satisfy $ζ_{p} \geq 1$ .

Statement references

For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges: convergence of a nonnegative nonincreasing series is equivalent to convergence of its condensed series.

For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges: a geometric series with ratio of absolute value less than one converges.

Integer powers $a^{m}$ : negative integer powers of a nonzero real are defined by reciprocals of natural powers.

The canonical natural $ι (n) = n \cdot 1_{F}$ of a field: positive natural indices are represented by their canonical real images.

Canonical naturals are positive and strictly increasing: canonical real images of positive naturals are positive.

Facts & Assumptions

Given: An integer $p \in {2, 3}$ and the sequence $a_{k} = k^{- p}$ for positive integer $k$ , with integer indices interpreted in $R$ by the canonical embedding.

[F1]

If a nonnegative sequence $(a_{k})_{k \geq 1}$ is nonincreasing, then $\sum_{k \geq 1} a_{k}$ converges if and only if $\sum_{r \geq 0} 2^{r} a_{2^{r}}$ converges (For a nonincreasing nonnegative sequence, $\sum a_{k}$ converges iff $\sum 2^{k} a_{2^{k}}$ converges).

[F2]

A geometric series $\sum_{r \geq 0} q^{r}$ converges whenever $∣ q ∣ < 1$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

[F3]

For a positive integer $k$ , its real image is positive, and its negative integer powers are reciprocals of its natural powers (Integer powers $a^{m}$ , The canonical natural $ι (n) = n \cdot 1_{F}$ of a field, Canonical naturals are positive and strictly increasing).

Proof

1.1givenF3algebra

For $p = 2$ or $p = 3$ , every $a_{k} = k^{- p}$ is positive. If $1 \leq k \leq ℓ$ , then $ℓ^{p} \geq k^{p} > 0$ : for $p = 2$ this follows from $ℓ^{2} - k^{2} = (ℓ - k) (ℓ + k) \geq 0$ , and for $p = 3$ from $ℓ^{3} - k^{3} = (ℓ - k) (ℓ^{2} + ℓ k + k^{2}) \geq 0$ . Taking reciprocals reverses the inequality, so $a_{ℓ} \leq a_{k}$ .

1.2givenF2F3inductionalgebra

For $p = 2$ , the condensed term is $2^{r} / (2^{r})^{2} = 1 / 2^{r} = (1 / 2)^{r}$ . For $p = 3$ , it is $2^{r} / (2^{r})^{3} = 1 / (2^{r})^{2} = (1 / 4)^{r}$ . The final power identities follow by induction on $r$ from the recursive definition of natural powers. Thus the condensed series is geometric with ratio $1 / 2$ or $1 / 4$ , respectively; [F2] makes both series converge.

2.1givenF1step 1.1step 1.2algebra∎

The sequence is nonnegative and nonincreasing by step 1.1, and its condensed series converges by step 1.2. Apply [F1] to conclude that $ζ_{p}$ converges. Its first term is $1$ and all terms are positive, hence $ζ_{p} \geq 1$ .

Depends on

Used by

A finite-mean polynomial-tail reward is not sub-Gaussian Example

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Sources

Stephen Semmes, Elements of Analysis (standard reference, not scraped)