The Weierstrass M-Test

From Numerical Bounds to Uniform Convergence

For a series of real numbers, comparison with a known convergent series is a standard way to establish convergence. A similar idea applies to series of functions, but uniform convergence requires one error bound that works for every input in the domain. The Weierstrass M-Test supplies such a bound: if the size of the $n$th function is controlled everywhere by a term $M_n$, and the numerical series $\sum M_n$ converges, then the function series converges uniformly.

Let $E$ be a nonempty set and let $f_n:E\to\mathbb{R}$. The function series $\sum_{n=1}^{\infty}f_n(x)$ is understood through its partial sums $S_N(x)=\sum_{n=1}^{N}f_n(x)$. Uniform convergence means that these partial sums converge uniformly on $E$. The useful feature of the M-Test is that its hypotheses involve only a bound on each term and an ordinary numerical series; they do not require finding the sum function in advance.

Theorem (Weierstrass M-Test): Let $E$ be a nonempty set, and let $f_n:E\to\mathbb{R}$. Suppose there are nonnegative real numbers $M_n$ such that $$ |f_n(x)|\leq M_n \qquad\text{for every }n\geq1\text{ and every }x\in E, $$ and suppose the numerical series $\sum_{n=1}^{\infty}M_n$ converges. Then $\sum_{n=1}^{\infty}f_n(x)$ converges absolutely for every $x\in E$, and its partial sums converge uniformly on $E$. In particular, the sum function is well defined. The proof is developed in detail in the next tutorial.

There are two distinct conclusions. At any fixed $x$, the comparison $|f_n(x)|\leq M_n$ gives absolute convergence of the numerical series $\sum f_n(x)$. Uniform convergence is stronger: the same numerical tail $\sum_{n>N}M_n$ controls the error for all $x\in E$ at once. This is why the majorants must not depend on $x$.

Finite Blocks and Uniformly Small Terms

Even before considering an infinite sum, the majorants control every finite block of terms. This elementary estimate is often the calculation that makes an application of the M-Test transparent.

Proposition (Finite-Block Bound): Suppose $|f_n(x)|\leq M_n$ for every $x\in E$, where each $M_n\geq0$. For integers $1\leq p\leq q$, $$ \sup_{x\in E}\left|\sum_{n=p}^{q}f_n(x)\right| \leq \sum_{n=p}^{q}M_n. $$

Proof. Fix any $x\in E$. The triangle inequality and the majorant bounds give $$ \left|\sum_{n=p}^{q}f_n(x)\right| \leq\sum_{n=p}^{q}|f_n(x)| \leq\sum_{n=p}^{q}M_n. $$ The right-hand side does not depend on $x$, so taking the supremum over $E$ preserves the inequality. $\square$

A convergent numerical series of nonnegative terms also forces its individual terms to tend to zero. The corresponding function consequence is uniform, because the same $M_n$ bounds every value of $f_n$.

Proposition (Summable Majorants Force Uniformly Vanishing Terms): Suppose $M_n\geq0$, $\sum_{n=1}^{\infty}M_n$ converges, and $|f_n(x)|\leq M_n$ for every $n$ and $x\in E$. Then $f_n\to0$ uniformly on $E$.

Proof. Convergence of $\sum M_n$ implies $M_n\to0$: if its partial sums converge, then $M_n=\sum_{k=1}^{n}M_k-\sum_{k=1}^{n-1}M_k\to0$. For every $n$, $$ \sup_{x\in E}|f_n(x)|\leq M_n. $$ Since $M_n\to0$, the supremum on the left tends to zero. By the supremum criterion for uniform convergence, this says precisely that $f_n$ converges uniformly to the zero function. $\square$

This proposition is a useful quick check, but it is not a substitute for the M-Test: knowing that the individual terms tend uniformly to zero does not by itself ensure that their series converges uniformly. The M-Test requires the stronger information that the bounds are summable.

Worked Applications

Worked Example: A Power Series on a Closed Short Interval

Consider $$ \sum_{n=1}^{\infty}\frac{x^n}{n(n+1)} \qquad\text{on }[0,1]. $$ For $x\in[0,1]$, we have $0\leq x^n\leq1$, and therefore $$ \left|\frac{x^n}{n(n+1)}\right| \leq\frac{1}{n(n+1)}. $$ The bounds form a convergent series because $$ \frac{1}{n(n+1)}=\frac1n-\frac1{n+1}, \qquad \sum_{n=1}^{N}\frac{1}{n(n+1)}=1-\frac{1}{N+1}\longrightarrow1. $$ The Weierstrass M-Test gives uniform and absolute convergence on $[0,1]$.

It also gives a direct estimate for truncation. For every $x\in[0,1]$, $$ \left|\sum_{n=N+1}^{\infty}\frac{x^n}{n(n+1)}\right| \leq\sum_{n=N+1}^{\infty}\frac{1}{n(n+1)} =\frac{1}{N+1}. $$ The bound does not depend on $x$, and tends to zero as $N\to\infty$. At the endpoint $x=1$, the sum is $1$, as the same telescoping calculation shows.

Worked Example: A Trigonometric Series on the Whole Real Line

Define $$ f_n(x)=\frac{\sin(nx)}{n(n+1)} \qquad (x\in\mathbb{R},\ n\geq1). $$ Since $|\sin(nx)|\leq1$ for every real $x$, $$ |f_n(x)|\leq\frac{1}{n(n+1)}. $$ The majorants sum to $1$, so the series $\sum_{n=1}^{\infty}f_n(x)$ converges uniformly and absolutely on all of $\mathbb{R}$. In particular, the domain need not be bounded for the M-Test to apply.

For example, after $N$ terms the error satisfies the uniform estimate $$ \sup_{x\in\mathbb{R}}\left|\sum_{n=N+1}^{\infty}\frac{\sin(nx)}{n(n+1)}\right| \leq\sum_{n=N+1}^{\infty}\frac{1}{n(n+1)} =\frac{1}{N+1}. $$ Each term is continuous, but the important point for uniform convergence is the global bound on its size. The estimate remains valid regardless of how large $x$ is.

Worked Example: A Rational Expression with a Global Majorant

For $x\in\mathbb{R}$, set $q(x)=x/(1+x^2)$ and consider $$ \sum_{n=1}^{\infty}\frac{q(x)^n}{n}. $$ First, $2|x|\leq1+x^2$, since $(|x|-1)^2\geq0$. Thus $|q(x)|\leq1/2$ for every real $x$. Consequently, $$ \left|\frac{q(x)^n}{n}\right| \leq\frac{1}{n2^n}. $$ The numerical series of bounds converges: $1/(n2^n)\leq1/2^n$, and the geometric series $\sum_{n=1}^{\infty}1/2^n$ converges. The M-Test therefore gives uniform absolute convergence on $\mathbb{R}$.

This example illustrates why it can be useful to simplify the expression inside a function term before selecting a majorant. The variable ranges over an unbounded set, but the particular expression $q(x)$ remains bounded. A bound based only on $|x|$ would not reveal the needed control.

Continuity of a Sum

Uniform convergence is useful not only because it controls approximation error. Combined with continuity of the individual terms, it also gives continuity of the sum. This is a direct application of the Uniform Limits of Continuous Functions Are Continuous theorem established earlier in the course.

Theorem (Continuity of a Series Sum under the M-Test): Let $I\subseteq\mathbb{R}$ be an interval. Suppose each $f_n:I\to\mathbb{R}$ is continuous and there are nonnegative numbers $M_n$ with $|f_n(x)|\leq M_n$ for every $n$ and $x\in I$. If $\sum_{n=1}^{\infty}M_n$ converges, then the function $$ f(x)=\sum_{n=1}^{\infty}f_n(x) $$ is continuous on $I$.

Proof. For each $N$, the partial sum $S_N=\sum_{n=1}^{N}f_n$ is continuous on $I$, because it is a finite sum of continuous functions. By the Weierstrass M-Test, $S_N$ converges uniformly on $I$ to $f$. The Uniform Limits of Continuous Functions Are Continuous theorem now implies that $f$ is continuous on $I$. $\square$

For the first worked example, each term $x^n/[n(n+1)]$ is continuous on $[0,1]$, so its sum is continuous there. In the trigonometric example, each term is continuous on $\mathbb{R}$, and the same theorem gives continuity of the sum on all of $\mathbb{R}$.

Choosing Bounds and Recognizing the Limitations

The crucial step in using the M-Test is finding constants $M_n$ that work simultaneously for every $x$ in the domain. A bound that depends on $x$ may prove pointwise convergence, but it does not provide the uniform control required by the test. Once a valid majorant is found, check that its numerical series really converges; bounds that merely tend to zero are not enough.

The test is sufficient, not necessary. A function series may converge uniformly even when a particular obvious choice of majorants is not summable, or when no useful summable majorants have been found. Failure to find an M-Test bound therefore does not establish failure of uniform convergence. Conversely, if the hypotheses do hold, the test supplies both uniform convergence and pointwise absolute convergence, whether or not the sum has a convenient closed form.

The finite-block estimate explains the mechanism behind the test: every block of function terms is controlled by the corresponding block of numerical bounds, with no dependence on the input. When the numerical tails become small, this control applies uniformly across the entire domain. The next tutorial supplies the full proof that turns this tail control into the M-Test’s uniform-convergence conclusion.

Takeaway: To apply the Weierstrass M-Test, bound each function term by a nonnegative number independent of the input, and verify that the numerical series of bounds converges. The resulting convergence is uniform, and continuous terms have a continuous sum.

Check Your Understanding

Use the hypotheses and applications of the M-Test to answer the following questions.

Which part of the M-Test hypothesis ensures that the bounds control every point of the domain?
Why does $\sum M_n<\infty$ imply that $f_n\to0$ uniformly when $|f_n(x)|\leq M_n$?
What uniform truncation-error bound follows for the series $\sum x^n/[n(n+1)]$ on $[0,1]$?
Why does the trigonometric example remain valid on the unbounded domain $\mathbb{R}$?
What additional hypothesis, beyond the M-Test bounds, lets us conclude that the sum function is continuous?

What You'll Learn