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

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The Weierstrass M-test gives absolute pointwise convergence and uniform convergence of a function series

Statement

Let $X$ be a set and let $f_{k} : X \to R$ . Suppose there is a sequence of nonnegative reals $(M_{k})$ such that

$∣ f_{k} (x) ∣ \leq M_{k} (k \in N, x \in X)$

and the scalar series $\sum M_{k}$ converges. Then $\sum f_{k} (x)$ converges absolutely for every $x \in X$ , and the function series $\sum f_{k}$ converges uniformly on $X$ .

Facts & Assumptions

Given: Functions $f_{k} : X \to R$ and nonnegative reals $M_{k}$ with $∣ f_{k} (x) ∣ \leq M_{k}$ for all $k, x$ , such that $\sum M_{k}$ converges.

[L1]

If $0 \leq a_{k} \leq b_{k}$ eventually and $\sum b_{k}$ converges, then $\sum a_{k}$ converges (If $0 \leq a_{k} \leq b_{k}$ eventually, convergence of $\sum b_{k}$ gives convergence of $\sum a_{k}$ , and divergence of $\sum a_{k}$ gives divergence of $\sum b_{k}$ ).

[L2]

A convergent real series has uniformly small scalar tails: for every real $ε > 0$ there is $N$ such that $∣ \sum_{k = m + 1}^{n} a_{k} ∣ < ε$ whenever $n > m \geq N$ (A series converges iff for every $ε > 0$ there is $N$ with $∣ a_{m + 1} + \dots + a_{n} ∣ < ε$ for all $n > m \geq N$ ).

[L3]

Repeated triangle inequalities for finite sums give $∣ \sum_{k = p}^{q} u_{k} ∣ \leq \sum_{k = p}^{q} ∣ u_{k} ∣$ , and finite sums preserve termwise inequalities (The triangle inequality, Basic properties of the absolute value, Finite sums and finite products, by recursion, Laws of finite sums and finite products).

[L4]

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, A series of real-valued functions and its pointwise and uniform convergence through its partial sums).

Proof

technique · direct

1.1

Fix $x \in X$ . Since $0 \leq ∣ f_{k} (x) ∣ \leq M_{k}$ for every $k$ , [L1] shows that $\sum ∣ f_{k} (x) ∣$ converges.

givenL1

1.2

Let $ε > 0$ . By [L2] choose $N$ such that $\sum_{k = m + 1}^{n} M_{k} < ε$ whenever $n > m \geq N$ , the absolute value being unnecessary because the terms are nonnegative.

L2choose

2.1

For $n > m \geq N$ and $x \in X$ , one has $∣ \sum_{k = m + 1}^{n} f_{k} (x) ∣ \leq \sum_{k = m + 1}^{n} ∣ f_{k} (x) ∣ \leq \sum_{k = m + 1}^{n} M_{k} < ε$ .

step 1.2L3

3.1

Step 1.1 gives absolute pointwise convergence, while step 2.1 and [L4] give uniform convergence of $\sum f_{k}$ .

step 1.1step 2.1L4∎

Depends on

Used by

Dependency tree · two levels

34 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

MIT OpenCourseWare 18.100B, Real Analysis, Lectures 20–21 (standard reference, not scraped)
University of Alberta Math 317, Infinite Series of Functions (standard reference, not scraped)