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

1.1givenL1

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

1.2L2choose

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.

2.1step 1.2L3

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

3.1step 1.1step 2.1L4∎

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

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

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources