Weierstrass M-Test [dark version]

Q: What two conditions on $M_k$ make the M-test work?

The constants $M_k$ must be nonnegative, and the numerical series $\sum_{k=1}^\infty M_k$ must converge. In addition, every function term must satisfy the uniform bound $|f_k(x)|\le M_k$ for all $x\in D$ and all $k$. Together these ensure the tails of $\sum f_k(x)$ can be controlled by the tails of $\sum M_k$.

Q: Why does pointwise convergence upgrade to uniform convergence under the M-test?

Because the inequality $|f_k(x)|\le M_k$ holds for every $x\in D$, the same tail estimate works simultaneously across the whole domain. Using the triangle inequality, the tail $\sum_{k=m}^n f_k(x)$ is bounded in absolute value by $\sum_{k=m}^n M_k$, and the supremum over $x$ inherits that bound. Since $\sum M_k$ has tails that go to zero, the supremum of the tails goes to zero too—exactly uniform convergence.

Q: How does the Cauchy/Koshi criterion appear in the proof?

Convergence of $\sum M_k$ implies its partial sums form a Cauchy sequence: for any $\varepsilon>0$, there exists $N$ such that for all $m,n\ge N$, $\sum_{k=m}^n M_k<\varepsilon$. The proof then applies $|f_k(x)|\le M_k$ and the triangle inequality to show $\left|\sum_{k=m}^n f_k(x)\right|\le \sum_{k=m}^n M_k<\varepsilon$ for every $x$, and therefore $\sup_{x\in D}\left|\sum_{k=m}^n f_k(x)\right|<\varepsilon$.

Q: What does the supremum norm statement mean for uniform convergence?

Uniform convergence can be expressed as: the supremum over $x\in D$ of the tail $\sum_{k=n+1}^\infty f_k(x)$ goes to zero as $n\to\infty$. Equivalently, for any $\varepsilon>0$, there is $N$ such that for all $m>n\ge N$, $\sup_{x\in D}\left|\sum_{k=n+1}^m f_k(x)\right|<\varepsilon$.

Q: How does the example $f_k(x)=\cos(kx)/k^2$ satisfy the M-test?

For all real $x$, $|\cos(kx)|\le 1$, so $|f_k(x)|=|\cos(kx)|/k^2\le 1/k^2$. Since $\sum_{k=1}^\infty 1/k^2$ converges, the M-test applies with $M_k=1/k^2$. The result is uniform convergence of $\sum_{k=1}^\infty \cos(kx)/k^2$ on $\mathbb{R}$.

TL;DR

Weierstrass’ M-test proves uniform convergence of $\sum f_{k} (x)$ by finding nonnegative constants $M_{k}$ that dominate $∣ f_{k} (x) ∣$ for every $x \in D$ .

Briefing Cornell Notes

Briefing

Weierstrass’ M-test gives a clean route to proving that a series of functions converges uniformly—by bounding every term with a single, summable “majorant” sequence. The payoff is practical: once uniform convergence is secured, the limit function is well-defined on the whole domain, and many operations (like exchanging limits with other processes) become legitimate.

Consider a sequence of functions $f_{k}$ defined on a common set $D$ , and form the series $\sum_{k = 1}^{\infty} f_{k} (x)$ . The M-test starts with two requirements. First, there must exist nonnegative constants $M_{k}$ such that the numerical series $\sum_{k = 1}^{\infty} M_{k}$ converges (so it does not diverge to infinity). Second, every function term must be uniformly dominated: for all $k \in N$ and all $x \in D$ , $∣ f_{k} (x) ∣ \leq M_{k} .$ These $M_{k}$ act as a uniform majorant for the entire family of functions.

From there, the comparison test for series guarantees convergence pointwise for each $x \in D$ . But the M-test goes further: because the bounds are uniform in $x$ , the convergence of the partial sums $\sum_{k = 1}^{n} f_{k} (x)$ is uniform as well. In standard terms, the partial sums converge in the supremum norm: the maximum (over $x \in D$ ) of the tail $\sum_{k = n + 1}^{\infty} f_{k} (x)$ can be made arbitrarily small by choosing $n$ large enough.

The proof uses the Cauchy criterion for series (often presented as a “Cauchy/Koshi criterion”). Since $\sum M_{k}$ converges, its tails $\sum_{k = m}^{n} M_{k}$ become arbitrarily small. The uniform bound $∣ f_{k} (x) ∣ \leq M_{k}$ then lets the same tail estimate transfer to $\sum_{k = m}^{n} f_{k} (x)$ : for any $ε > 0$ , one can pick an index $N$ so that whenever $m, n \geq N$ , $x \in D sup k = m \sum n f_{k} (x) \leq ε .$ That supremum-tail control is exactly what uniform convergence means.

A concrete application makes the criterion feel immediate. Take $f_{k} (x) = \frac{c o s ( k x )}{k ^{2}}$ on $R$ . Since $∣ cos (k x) ∣ \leq 1$ , every term satisfies $∣ f_{k} (x) ∣ \leq 1/ k^{2}$ for all $x$ . The series $\sum 1/ k^{2}$ converges, so the M-test concludes that $\sum_{k = 1}^{\infty} \frac{c o s ( k x )}{k ^{2}}$ converges uniformly on $R$ . The uniform convergence is especially useful when differentiating term-by-term in related Fourier/series settings, where controlling tails uniformly helps justify exchanging limits and derivatives.

Cornell Notes

Weierstrass’ M-test provides a sufficient condition for uniform convergence of a function series $\sum_{k = 1}^{\infty} f_{k} (x)$ on a domain $D$ . If there exist nonnegative constants $M_{k}$ such that $\sum M_{k}$ converges and $∣ f_{k} (x) ∣ \leq M_{k}$ for every $x \in D$ and every $k$ , then the series converges uniformly (and the series of absolute values also behaves well). The proof transfers the Cauchy (Koshi) tail smallness from $\sum M_{k}$ to the tails of $\sum f_{k} (x)$ using the triangle inequality and the supremum norm. A key example is $f_{k} (x) = cos (k x) / k^{2}$ , where $∣ f_{k} (x) ∣ \leq 1/ k^{2}$ and $\sum 1/ k^{2}$ converges, yielding uniform convergence on $R$ .

What two conditions on $M_{k}$ make the M-test work?

The constants

M_{k}

must be nonnegative, and the numerical series

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

must converge. In addition, every function term must satisfy the uniform bound

∣ f_{k} (x) ∣ \leq M_{k}

for all

x \in D

and all

k

. Together these ensure the tails of

\sum f_{k} (x)

can be controlled by the tails of

\sum M_{k}

Why does pointwise convergence upgrade to uniform convergence under the M-test?

Because the inequality

∣ f_{k} (x) ∣ \leq M_{k}

holds for every

x \in D

, the same tail estimate works simultaneously across the whole domain. Using the triangle inequality, the tail

\sum_{k = m}^{n} f_{k} (x)

is bounded in absolute value by

\sum_{k = m}^{n} M_{k}

, and the supremum over

x

inherits that bound. Since

\sum M_{k}

has tails that go to zero, the supremum of the tails goes to zero too—exactly uniform convergence.

How does the Cauchy/Koshi criterion appear in the proof?

Convergence of

\sum M_{k}

implies its partial sums form a Cauchy sequence: for any

ε > 0

, there exists

N

such that for all

m, n \geq N

\sum_{k = m}^{n} M_{k} < ε

. The proof then applies

∣ f_{k} (x) ∣ \leq M_{k}

and the triangle inequality to show

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

for every

x

, and therefore

sup_{x \in D} ∣ \sum_{k = m}^{n} f_{k} (x) ∣ < ε

What does the supremum norm statement mean for uniform convergence?

Uniform convergence can be expressed as: the supremum over

x \in D

of the tail

\sum_{k = n + 1}^{\infty} f_{k} (x)

goes to zero as

n \to \infty

. Equivalently, for any

ε > 0

, there is

N

such that for all

m > n \geq N

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

How does the example $f_{k} (x) = cos (k x) / k^{2}$ satisfy the M-test?

For all real

x

∣ cos (k x) ∣ \leq 1

, so

∣ f_{k} (x) ∣ = ∣ cos (k x) ∣/ k^{2} \leq 1/ k^{2}

. Since

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

converges, the M-test applies with

M_{k} = 1/ k^{2}

. The result is uniform convergence of

\sum_{k = 1}^{\infty} cos (k x) / k^{2}

R

Review Questions

Given $∣ f_{k} (x) ∣ \leq M_{k}$ for all $x \in D$ , what additional property of $\sum M_{k}$ is required to conclude uniform convergence?
How would you use the triangle inequality to bound $∣ \sum_{k = m}^{n} f_{k} (x) ∣$ in terms of $M_{k}$ ?
For $f_{k} (x) = cos (k x) / k^{p}$ , for which values of $p$ would the M-test guarantee uniform convergence on $R$ ? (Assume the same reasoning as the $p = 2$ example.)

Key Points

1
Weierstrass’ M-test proves uniform convergence of $\sum f_{k} (x)$ by finding nonnegative constants $M_{k}$ that dominate $∣ f_{k} (x) ∣$ for every $x \in D$ .
2
Uniform domination requires $∣ f_{k} (x) ∣ \leq M_{k}$ for all $k$ and all $x \in D$ , not just pointwise bounds at individual points.
3
The numerical series $\sum M_{k}$ must converge; its tail smallness drives the uniform tail smallness of $\sum f_{k} (x)$ .
4
Once the supremum of the tails $sup_{x \in D} ∣ \sum_{k = m}^{n} f_{k} (x) ∣$ goes to zero, the partial sums converge uniformly in the supremum norm.
5
The M-test also yields strong control over the absolute values, since the same majorant bounds $∣ f_{k} (x) ∣$ .
6
A standard application is $\sum cos (k x) / k^{2}$ , where $∣ cos (k x) ∣ \leq 1$ and $\sum 1/ k^{2}$ converges, giving uniform convergence on $R$ .
7
Uniform convergence is particularly valuable when justifying operations like term-by-term differentiation in series-based analysis.

Highlights

Uniform convergence follows once every term satisfies a global bound ∣fk​(x)∣≤Mk​ and the majorant series ∑Mk​ converges.

The proof transfers the Cauchy tail estimate from ∑Mk​ to ∑fk​(x) using the triangle inequality and then takes a supremum over D.

For fk​(x)=cos(kx)/k2, the simple inequality ∣cos(kx)∣≤1 reduces the problem to the convergence of ∑1/k2.

Topics

Weierstrass M-Test
Uniform Convergence
Function Series
Cauchy Criterion
Majorant Series