Weierstrass M-Test

Q: What two conditions make the Weierstrass M-test work for $\sum f_k(x)$?

First, there must be nonnegative constants $M_k\ge 0$ such that $|f_k(x)|\le M_k$ for every $k\in\mathbb{N}$ and every $x\in D$. Second, the numerical series $\sum_{k=1}^\infty M_k$ must converge (its partial sums approach a finite limit). Together, these ensure the function-series tails are uniformly small.

Q: How does the test connect uniform convergence to the supremum norm?

Uniform convergence means $\|S-S_n\|_\infty=\sup_{x\in D}|S(x)-S_n(x)|\to 0$. Using the domination $|f_k(x)|\le M_k$, the tail satisfies $|\sum_{k=n+1}^m f_k(x)|\le \sum_{k=n+1}^m M_k$. Since the right-hand side does not depend on $x$, taking $\sup_{x\in D}$ preserves the same $\varepsilon$-bound for all points at once.

Q: Why does convergence of $\sum M_k$ imply that $\sum f_k(x)$ is Cauchy uniformly?

Convergence of $\sum M_k$ gives the Cauchy criterion for series: for any $\varepsilon>0$, there exists $N$ so that for all $m>n\ge N$, $\sum_{k=n+1}^m M_k<\varepsilon$. For any $x\in D$, the tail of the function series is bounded by $\sum_{k=n+1}^m |f_k(x)|\le \sum_{k=n+1}^m M_k$, so $|\sum_{k=n+1}^m f_k(x)|<\varepsilon$ uniformly in $x$.

Q: What does the M-test guarantee about $\sum |f_k(x)|$?

Because the same bound works after taking absolute values—namely $|f_k(x)|\le M_k$—the numerical comparison also applies to the absolute-value series. As a result, $\sum |f_k(x)|$ converges uniformly on $D$ whenever $\sum M_k$ converges.

Q: How does the example $f_k(x)=\frac{\cos(kx)}{k^2}$ fit the M-test?

On $\mathbb{R}$, $|\cos(kx)|\le 1$ for all $x$. Therefore $|f_k(x)|=\left|\frac{\cos(kx)}{k^2}\right|\le \frac{1}{k^2}$. Since $\sum_{k=1}^\infty \frac{1}{k^2}$ converges, the M-test implies $\sum_{k=1}^\infty \frac{\cos(kx)}{k^2}$ converges uniformly on $\mathbb{R}$.

TL;DR

Uniform convergence of $\sum f_{k} (x)$ follows if $∣ f_{k} (x) ∣ \leq M_{k}$ for all $x \in D$ and if $\sum M_{k}$ converges.

Briefing Cornell Notes

Briefing

The Weierstrass M-test provides a clean, practical way to prove uniform convergence for series of functions. If a series of functions $\sum_{k = 1}^{\infty} f_{k} (x)$ can be dominated by a single convergent numerical series $\sum_{k = 1}^{\infty} M_{k}$ , then the function series converges uniformly on the whole domain—no point-by-point checking required. The payoff is substantial: uniform convergence is strong enough to guarantee that the limit function is well-defined and behaves predictably under operations like taking absolute values and (in many contexts) differentiating term-by-term.

Concretely, the test starts with a common domain $D$ for all functions $f_{k}$ , which may be real- or complex-valued. One assumes there exist constants $M_{k} \geq 0$ such that for every $k \in N$ and every $x \in D$ , the inequality $∣ f_{k} (x) ∣ \leq M_{k}$ holds. The second requirement is that the numerical series $\sum_{k = 1}^{\infty} M_{k}$ converges (so its partial sums approach a finite limit). Under these two conditions, the series $\sum f_{k} (x)$ converges for every $x \in D$ , and—crucially—it converges uniformly on $D$ .

Uniform convergence is framed using the supremum norm: if $S (x) = \sum_{k = 1}^{\infty} f_{k} (x)$ denotes the pointwise limit, then the partial sums $S_{n} (x) = \sum_{k = 1}^{n} f_{k} (x)$ satisfy $∥ S - S_{n} ∥_{\infty} = sup_{x \in D} ∣ S (x) - S_{n} (x) ∣ \to 0$ as $n \to \infty$ . Intuitively, the graphs of the partial sums eventually stay inside any chosen “ $ε$ -tube” around the graph of $S$ , uniformly across all $x \in D$ .

The proof relies on the Cauchy (Cauchy–Koshi) criterion for series, using completeness of $R$ or $C$ . Since $\sum M_{k}$ converges, its tails become arbitrarily small: for any $ε > 0$ , there exists $N$ such that for all $m > n \geq N$ , $\sum_{k = n + 1}^{m} M_{k} < ε$ . For any fixed $x \in D$ , the tail of the function series satisfies $∣ \sum_{k = n + 1}^{m} f_{k} (x) ∣ \leq \sum_{k = n + 1}^{m} ∣ f_{k} (x) ∣ \leq \sum_{k = n + 1}^{m} M_{k}$ , so the same $ε$ -control holds uniformly. Taking the supremum over $x$ preserves the bound, yielding $∥ S - S_{n} ∥_{\infty} \to 0$ . The test also applies to $\sum ∣ f_{k} (x) ∣$ , giving uniform convergence of the absolute-value series as well.

A standard application illustrates the method: let $f_{k} (x) = \frac{c o s ( k x )}{k ^{2}}$ on $R$ . Since $∣ cos (k x) ∣ \leq 1$ , one has $∣ f_{k} (x) ∣ \leq 1/ k^{2}$ . The numerical series $\sum 1/ k^{2}$ converges, so $\sum \frac{c o s ( k x )}{k ^{2}}$ converges uniformly on $R$ . That uniform convergence is especially useful in Fourier-related settings, including justifying operations like differentiation of related series.

Cornell Notes

The Weierstrass M-test gives a sufficient condition for uniform convergence of a function series $\sum_{k = 1}^{\infty} f_{k} (x)$ on a domain $D$ . If there are constants $M_{k} \geq 0$ with $∣ f_{k} (x) ∣ \leq M_{k}$ for every $x \in D$ and if the numerical series $\sum M_{k}$ converges, then $\sum f_{k} (x)$ converges uniformly on $D$ . The limit function $S (x) = \sum f_{k} (x)$ is well-defined, and the tail $S (x) - S_{n} (x)$ becomes small in the supremum norm: $sup_{x \in D} ∣ S (x) - S_{n} (x) ∣ \to 0$ . The same domination also yields uniform convergence of $\sum ∣ f_{k} (x) ∣$ .

What two conditions make the Weierstrass M-test work for $\sum f_{k} (x)$ ?

First, there must be nonnegative constants

M_{k} \geq 0

such that

∣ f_{k} (x) ∣ \leq M_{k}

for every

k \in N

and every

x \in D

. Second, the numerical series

\sum_{k = 1}^{\infty} M_{k}

must converge (its partial sums approach a finite limit). Together, these ensure the function-series tails are uniformly small.

How does the test connect uniform convergence to the supremum norm?

Uniform convergence means

∥ S - S_{n} ∥_{\infty} = sup_{x \in D} ∣ S (x) - S_{n} (x) ∣ \to 0

. Using the domination

∣ f_{k} (x) ∣ \leq M_{k}

, the tail satisfies

∣ \sum_{k = n + 1}^{m} f_{k} (x) ∣ \leq \sum_{k = n + 1}^{m} M_{k}

. Since the right-hand side does not depend on

x

, taking

sup_{x \in D}

preserves the same

ε

-bound for all points at once.

Why does convergence of $\sum M_{k}$ imply that $\sum f_{k} (x)$ is Cauchy uniformly?

Convergence of

\sum M_{k}

gives the Cauchy criterion for series: for any

ε > 0

, there exists

N

so that for all

m > n \geq N

\sum_{k = n + 1}^{m} M_{k} < ε

. For any

x \in D

, the tail of the function series is bounded by

\sum_{k = n + 1}^{m} ∣ f_{k} (x) ∣ \leq \sum_{k = n + 1}^{m} M_{k}

, so

∣ \sum_{k = n + 1}^{m} f_{k} (x) ∣ < ε

uniformly in

x

What does the M-test guarantee about $\sum ∣ f_{k} (x) ∣$ ?

Because the same bound works after taking absolute values—namely

∣ f_{k} (x) ∣ \leq M_{k}

—the numerical comparison also applies to the absolute-value series. As a result,

\sum ∣ f_{k} (x) ∣

converges uniformly on

D

whenever

\sum M_{k}

converges.

How does the example $f_{k} (x) = \frac{c o s ( k x )}{k ^{2}}$ fit the M-test?

R

∣ cos (k x) ∣ \leq 1

for all

x

. Therefore

∣ f_{k} (x) ∣ = \frac{c o s ( k x )}{k ^{2}} \leq \frac{1}{k ^{2}}

. Since

\sum_{k = 1}^{\infty} \frac{1}{k ^{2}}

converges, the M-test implies

\sum_{k = 1}^{\infty} \frac{c o s ( k x )}{k ^{2}}

converges uniformly on

R

Review Questions

State the Weierstrass M-test precisely, including the role of $M_{k}$ and the convergence requirement.
Explain why the inequality $∣ f_{k} (x) ∣ \leq M_{k}$ leads to a bound on $∣ S_{m} (x) - S_{n} (x) ∣$ that is independent of $x$ .
Given $∣ f_{k} (x) ∣ \leq M_{k}$ on $D$ , what additional information about $\sum M_{k}$ is needed to conclude uniform convergence of $\sum f_{k} (x)$ ?

Key Points

1
Uniform convergence of $\sum f_{k} (x)$ follows if $∣ f_{k} (x) ∣ \leq M_{k}$ for all $x \in D$ and if $\sum M_{k}$ converges.
2
The supremum-norm criterion $sup_{x \in D} ∣ S (x) - S_{n} (x) ∣ \to 0$ is the uniform convergence target.
3
The proof uses the Cauchy (Cauchy–Koshi) criterion for series and the triangle inequality to control tails.
4
Because the domination is pointwise and independent of $x$ , the same $ε$ -tail bound holds uniformly across the entire domain.
5
The M-test also yields uniform convergence of $\sum ∣ f_{k} (x) ∣$ under the same assumptions.
6
For $f_{k} (x) = cos (k x) / k^{2}$ , the bound $∣ f_{k} (x) ∣ \leq 1/ k^{2}$ and convergence of $\sum 1/ k^{2}$ guarantee uniform convergence on $R$ .

Highlights

If a single convergent numerical series ∑Mk​ uniformly dominates ∣fk​(x)∣, then ∑fk​(x) converges uniformly on the whole domain.

Uniform convergence is proved by bounding the tails ∣Sm​(x)−Sn​(x)∣ with ∑k=n+1m​Mk​, then taking a supremum over x.

The example ∑cos(kx)/k2 converges uniformly on R because ∣cos(kx)∣≤1 and ∑1/k2 converges.

Topics

Weierstrass M-Test
Uniform Convergence
Series of Functions
Cauchy Criterion
Supremum Norm

Briefing

Cornell Notes

What two conditions make the Weierstrass M-test work for ∑fk​(x)?

How does the test connect uniform convergence to the supremum norm?

Why does convergence of ∑Mk​ imply that ∑fk​(x) is Cauchy uniformly?

What does the M-test guarantee about ∑∣fk​(x)∣?

How does the example fk​(x)=k2cos(kx)​ fit the M-test?