LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-07-31

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If $ι (n + 1) a_{n} \to 0$ , short multiplicative blocks of the coefficients have uniformly small sums

Statement

Suppose $ι (n + 1) a_{n} \to 0$ . For every $ε > 0$ there is $N_{0}$ such that, whenever $q \geq p \geq N_{0}$ ,

$\sum_{n = p}^{q} ∣ a_{n} ∣ \leq ε \frac{ι (q - p + 1)}{ι (p + 1)} .$

Moreover, for $N \geq N_{0}$ and $x_{N} : = 1 - 1 / ι (N + 1)$ ,

$\sum_{n = N_{0}}^{N} ∣ a_{n} ∣ (1 - x_{N}^{n}) \leq ε, \sum_{n > N} ∣ a_{n} ∣ x_{N}^{n} \leq ε .$

Facts & Assumptions

Given: The Tauber condition $ι (n + 1) a_{n} \to 0$ .

[L1]

The canonical naturals $ι (n + 1)$ are positive and strictly increasing, positive reciprocals reverse order, and $∣ u v ∣ = ∣ u ∣ ∣ v ∣$ (Canonical naturals are positive and strictly increasing, Inverses of positives are positive, and reciprocation reverses order, Basic properties of the absolute value).

[L2]

Real-sequence convergence is tested with positive rational tolerances, and below every positive real lies a positive rational (Limits and Cauchy sequences of reals, The rationals embed densely in the reals).

[L3]

For $0 \leq x \leq 1$ , multiplying out the finite sum gives $1 - x^{n} = (1 - x) \sum_{k = 0}^{n - 1} x^{k} \leq ι (n) (1 - x)$ .

[L4]

For $0 \leq x < 1$ , the geometric-series formula gives $\sum_{n > N} x^{n} = x^{N + 1} / (1 - x)$ (For $∣ r ∣ < 1$ , $\sum_{k \geq 0} r^{k} = 1 / (1 - r)$ , and for $∣ r ∣ \geq 1$ the series diverges).

Proof

technique · direct

1.1

Choose a positive rational $δ < ε$ . By the limit hypothesis and [L2], choose $N_{0}$ so that $∣ ι (n + 1) a_{n} ∣ < δ$ for $n \geq N_{0}$ . Positivity and multiplicativity in [L1] give $∣ a_{n} ∣ < δ / ι (n + 1) < ε / ι (n + 1)$ there. Since $1 / ι (n + 1) \leq 1 / ι (p + 1)$ on $p \leq n \leq q$ , summing proves the block estimate.

L1L2choosealgebra

2.1

For $N_{0} \leq n \leq N$ , [L3] gives $∣ a_{n} ∣ (1 - x_{N}^{n}) \leq ε (1 - x_{N})$ . There are at most $N + 1$ terms and $ι (N + 1) (1 - x_{N}) = 1$ , proving the first weighted estimate.

step 1.1L3algebra

3.1

For $n > N$ , step 1.1 gives $∣ a_{n} ∣ \leq ε / ι (N + 2)$ . Summing the geometric tail yields $\sum_{n > N} ∣ a_{n} ∣ x_{N}^{n} \leq ε x_{N}^{N + 1} ι (N + 1) / ι (N + 2) \leq ε$ .

step 1.1L4algebra∎

Depends on

Used by

Tauber's theorem: an Abel-summable series with ι(n+1)aₙ→0 converges ordinarily to its Abel sum Theorem

Dependency tree · two levels

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Sources

Tauberian theorem, Encyclopedia of Mathematics (standard reference, not scraped)
Tauberian theorems, Encyclopedia of Mathematics (standard reference, not scraped)

If ι(n+1)an→0ι(n+1)an​→0, short multiplicative blocks of the coefficients have uniformly small sums