LemmaStatement: AI-adaptedProof: AI-generatedPipeline-generatedjudge pass (gpt-5.6-terra)audited 2026-09-07

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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.

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Kronecker summation lemma

Statement

Let $(x_{n})_{n \geq 1}$ be real and let $0 < b_{n}$ be deterministic, nondecreasing, and tend to infinity. If $\sum_{n \geq 1} x_{n} / b_{n}$ converges in $R$ , then $\frac{1}{b_{n}} \sum_{k = 1}^{n} x_{k} ⟶ 0.$ Repeated values of $b_{n}$ are allowed.

Facts & Assumptions

[F1]

Abel summation by parts: with $A_{n} = \sum_{k < n} a_{k}$ one has $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k})$ for every $n \geq 1$ : Let $(a_{k})$ and $(b_{k})$ be sequences of reals and let $A_{n} : = \sum_{k < n} a_{k} (n \in N)$ be the partial sums of $\sum a_{k}$ (def-series, def-finite-sum), so that $A_{0} = 0$ and $a_{k} = A_{k + 1} - A_{k}$ for every $k$ . Then for every natural number $n \geq 1$ $\sum_{k < n} a_{k} b_{k} = A_{n} b_{n - 1} - \sum_{k < n - 1} A_{k + 1} (b_{k + 1} - b_{k}) .$ Both sides are finite sums in the sense of def-finite-sum; at $n = 1$ the right-hand sum is empty and the identity reads $a_{0} b_{0} = A_{1} b_{0}$ . The hypothesis $n \geq 1$ is what makes the statement legitimate, not merely convenient: the index $n - 1$ occurs on the right, and $n - 1$ is a natural number exactly when $n \geq 1$ . At $n = 0$ there is nothing to state, both the left-hand side and $A_{0}$ being $0$ .

Proof

Given: The objects and hypotheses of the statement.

1.1

Put $t_{0} = b_{0} = 0$ and $t_{n} = \sum_{k = 1}^{n} x_{k} / b_{k} \to t$ . Abel summation, shifted from its zero-based indices, gives $b_{n}^{- 1} \sum_{k = 1}^{n} x_{k} = t_{n} - \sum_{k = 1}^{n} (b_{k} - b_{k - 1}) t_{k - 1} / b_{n}$ . One can verify the same identity by substituting $x_{k} = b_{k} (t_{k} - t_{k - 1})$ and telescoping; for $n = 1$ it reads $x_{1} / b_{1} = t_{1}$ .

F1givenalgebra

2.1

The weights $w_{n, k} = (b_{k} - b_{k - 1}) / b_{n}$ are nonnegative and sum to one. For $ε > 0$ , take $M$ such that $∣ t_{j} - t ∣ < ε$ for $j \geq M$ . With $K = \max_{0 \leq j < M} ∣ t_{j} - t ∣$ , the weighted average differs from $t$ by at most $K b_{M} / b_{n} + ε$ for $n > M$ . The first term tends to zero; no division by $K$ is required, even if it vanishes. Repeated normalizers merely give zero weights.

step 1.1givenalgebra

3.1

As $ε$ is arbitrary the weighted average tends to $t$ . Subtracting it from $t_{n} \to t$ in the finite identity proves the assertion, including the zero sequence.

step 1.1step 2.1algebra∎

Depends on

Used by

Strong law under summable normalized variances Theorem

Dependency tree · two levels

10 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Theorem 2.5.9 and full proof, pp. 85–86 (standard reference, not scraped)