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

How statement and proof provenance work

The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

Uniform Dirichlet test: uniformly bounded partial sums times a uniformly decreasing null family give a uniformly convergent function series

Statement

Let $X$ be a set and let $u_{k}, v_{k} : X \to R$ . Put $A_{n} (x) : = \sum_{k < n} u_{k} (x)$ . Suppose:

there is $M \geq 0$ such that $∣ A_{n} (x) ∣ \leq M$ for every $n$ and $x$ ;
$v_{k} (x) \geq 0$ and $v_{k + 1} (x) \leq v_{k} (x)$ for every $k$ and $x$ ;
$v_{k} \to 0$ uniformly on $X$ .

Then the function series $\sum u_{k} v_{k}$ converges uniformly on $X$ .

Facts & Assumptions

Given: Functions $u_{k}, v_{k} : X \to R$ satisfying the three hypotheses in the Statement, with partial sums $A_{n} (x) = \sum_{k < n} u_{k} (x)$ .

[L1]

Abel summation by parts expresses a finite sum $\sum_{j < r} a_{j} b_{j}$ as $A_{r} b_{r - 1} - \sum_{j < r - 1} A_{j + 1} (b_{j + 1} - b_{j})$ , where $A_{r} = \sum_{j < r} a_{j}$ (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 $n \geq 1$ ).

[L2]

Finite sums split and telescope, preserve inequalities, and obey the triangle inequality after repeated use of $∣ s + t ∣ \leq ∣ s ∣ + ∣ t ∣$ (Finite sums and finite products, by recursion, Laws of finite sums and finite products, The triangle inequality, Basic properties of the absolute value).

[L3]

A function series converges uniformly exactly when its tails are uniformly small (A series of real-valued functions converges uniformly if and only if its tails are uniformly small).

Proof

technique · direct

1.1

Let $ε > 0$ . Uniform convergence $v_{k} \to 0$ gives $N$ such that $0 \leq v_{k} (x) < ε / (2 M + 1)$ for every $k \geq N$ and every $x \in X$ .

givenchoose

1.2

Fix $n > m \geq N$ and $x \in X$ , put $p : = m + 1$ , $q : = n$ , and for $p \leq r \leq q$ put $B_{r} (x) : = \sum_{k = p}^{r} u_{k} (x) = A_{r + 1} (x) - A_{p} (x)$ .

L2construct

2.1

For every $p \leq r \leq q$ , the bound on the $A_{j}$ gives $∣ B_{r} (x) ∣ \leq ∣ A_{r + 1} (x) ∣ + ∣ A_{p} (x) ∣ \leq 2 M$ .

step 1.2L2

2.2

Applying [L1] to the shifted finite list from $p$ through $q$ gives $\sum_{k = p}^{q} u_{k} (x) v_{k} (x) = B_{q} (x) v_{q} (x) + \sum_{k = p}^{q - 1} B_{k} (x) (v_{k} (x) - v_{k + 1} (x))$ .

step 1.2L1L2

3.1

Since $v_{k} (x) - v_{k + 1} (x) \geq 0$ , steps 2.1 and 2.2 with telescoping give $∣ \sum_{k = p}^{q} u_{k} (x) v_{k} (x) ∣ \leq 2 M v_{q} (x) + 2 M \sum_{k = p}^{q - 1} (v_{k} (x) - v_{k + 1} (x)) = 2 M v_{p} (x) < ε$ .

step 1.1step 2.1step 2.2L2algebra

4.1

The estimate in step 3.1 holds for every $n > m \geq N$ and $x \in X$ , so [L3] proves uniform convergence of $\sum u_{k} v_{k}$ .

step 3.1L3∎

Depends on

Used by

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Sources

University of Alberta Math 317, Infinite Series of Functions (standard reference, not scraped)