The Weierstrass M-test

Statement

Let \( E \) be a non-empty set and let \( (f_n)_{n \ge 1} \) be a sequence of functions \( f_n : E \to \mathbb{R} \) (or \( \mathbb{C} \)). Suppose there exists a sequence of real constants \( (M_n)_{n \ge 1} \), independent of \( x \), such that (i) \( |f_n(x)| \le M_n \) for every \( x \in E \) and every \( n \in \mathbb{N} \), and (ii) the numerical series \( \sum_{n=1}^{\infty} M_n \) converges. Then the series \( \sum_{n=1}^{\infty} f_n \) converges absolutely at every point of \( E \), and the sequence of partial sums \( S_N = \sum_{n=1}^{N} f_n \) converges uniformly on \( E \) to the function \( f(x) = \sum_{n=1}^{\infty} f_n(x) \).

Why it matters

Uniform convergence is the currency that lets limit operations pass through an infinite sum: a uniform limit of continuous functions is continuous, a uniformly convergent series may be integrated term by term, and (with a companion hypothesis) differentiated term by term. But verifying uniform convergence from the definition means estimating \( \sup_{x \in E} |f(x) - S_N(x)| \), which requires already knowing the sum \( f \). The M-test removes this obstacle entirely: it converts a question about functions into a question about a single series of non-negative numbers, checkable by the ordinary convergence tests of first-year analysis.

It is the workhorse behind the basic theory of power series, Fourier series, the Riemann zeta function on half-planes, and Weierstrass's own construction of a continuous nowhere-differentiable function. Structurally it is the prototype of a deep principle: in a complete space, absolute convergence implies convergence.

Hypotheses

Proof

Result

Corollaries & converses

• Continuity of the sum. If moreover \( E \) is a metric (or topological) space and each \( f_n \) is continuous, then \( f = \sum f_n \) is continuous, by the uniform limit theorem. This is the single most-used consequence.
• Term-by-term integration. If \( E = [a,b] \) and each \( f_n \) is Riemann integrable, then \( f \) is Riemann integrable and \( \int_a^b f = \sum_{n=1}^{\infty} \int_a^b f_n \), since uniform convergence permits the interchange of limit and integral on a bounded interval.
• Power series. A power series \( \sum a_n x^n \) with radius of convergence \( R \) converges uniformly on every closed interval \( [-r, r] \) with \( 0 \lt r \lt R \) (take \( M_n = |a_n| r^n \)); hence its sum is continuous on \( (-R, R) \). The same argument in \( \mathbb{C} \) gives uniform convergence on closed subdiscs.
• Converse fails. Uniform convergence does not imply the existence of summable dominating constants. On \( [0, \infty) \), the series \( \sum_{n=1}^{\infty} \frac{(-1)^n}{n + x} \) converges uniformly (alternating-series remainder \( \le \frac{1}{N+1} \), independent of \( x \)), yet any admissible \( M_n \) must satisfy \( M_n \ge \sup_{x \ge 0} \frac{1}{n+x} = \frac{1}{n} \), so \( \sum M_n \) diverges. The M-test also forces absolute convergence, which this series lacks — a second, structural reason no converse can hold.
• Abstract form. The M-test is exactly the statement that the space \( B(E) \) of bounded functions on \( E \) with the sup norm is a Banach space, applied to an absolutely convergent series: \( \sum \lVert f_n \rVert_{\infty} \le \sum M_n \lt \infty \) implies norm convergence of \( \sum f_n \).

Fails without

Common errors

• Letting \( M_n \) depend on \( x \). Writing \( |x^n| \le x^n \) and "summing" is circular; the test requires one constant per term, valid across the entire domain. Always compute or bound \( \sup_{x \in E} |f_n(x)| \) first.
• Reading the test as an equivalence. Concluding "the M-test fails, therefore the convergence is not uniform." The test is sufficient, not necessary — see \( \sum \frac{(-1)^n}{n+x} \) above. Failure of the M-test proves nothing.
• Forgetting to check summability. Finding \( M_n = \sup |f_n| \) correctly and then never testing \( \sum M_n \). E.g. \( \sup_{x \in [0,1]} x^n (1-x) = \frac{1}{n+1} \left( \frac{n}{n+1} \right)^n \sim \frac{1}{e\,n} \), which is not summable.
• Ignoring the domain. \( \sum \frac{x^n}{n} \) passes the M-test on \( [-r, r] \) for each \( r \lt 1 \) but on no neighbourhood of \( 1 \); "uniform on every compact subset" and "uniform on the whole open interval" are different conclusions, and only the former follows.
• Differentiating term by term on the strength of the M-test alone. Uniform convergence of \( \sum f_n \) says nothing about \( \sum f_n' \); the term-by-term differentiation theorem needs uniform convergence of the derived series (plus convergence of the original at one point). The Weierstrass function is the canonical warning.

Discussion

The test is due to Karl Weierstrass, who used it in his Berlin lectures from the 1860s onward as part of his programme to rebuild analysis on \( \varepsilon \)–\( \delta \) foundations. The historical backdrop is Cauchy's famous 1821 misstep — the claim that a convergent series of continuous functions has a continuous sum — refuted by Abel with Fourier-type series such as \( \sum \frac{\sin nx}{n} \), whose sum jumps. The repair required isolating the notion of uniform convergence (Seidel, Stokes, and above all Weierstrass), and the M-test was the tool that made the notion usable: it is often the only uniform-convergence criterion a working analyst ever needs.

Its most spectacular early application was Weierstrass's 1872 example \( W(x) = \sum_{n=0}^{\infty} a^n \cos(b^n \pi x) \) with \( 0 \lt a \lt 1 \): the M-test with \( M_n = a^n \) shows instantly that \( W \) is continuous on \( \mathbb{R} \), while a delicate separate argument (for \( ab \) large enough) shows it is differentiable nowhere. The test thus certified the existence of objects that shattered the era's geometric intuition — continuity with no tangent anywhere — and marks the point where analysis decisively outgrew pictures.

Within the standard curriculum the M-test is the engine behind the elementary theory of power series (uniform convergence on compact subdiscs, hence continuity, term-by-term integration, and — applied to the derived series, which has the same radius — term-by-term differentiation), behind the continuity of \( \zeta(s) = \sum n^{-s} \) on half-planes \( \operatorname{Re} s \ge 1 + \delta \), and behind the routine manipulation of Fourier series with summable coefficients.

The modern view absorbs the theorem into Banach space theory. For a normed space \( X \), the following are equivalent: (i) \( X \) is complete; (ii) every series in \( X \) with \( \sum \lVert x_n \rVert \lt \infty \) converges in \( X \). The M-test is precisely implication (i) \( \Rightarrow \) (ii) applied to \( X = B(E) \), the bounded functions on \( E \) under \( \lVert \cdot \rVert_{\infty} \), whose completeness is the uniform Cauchy criterion in disguise; when \( E \) is a compact metric space one may take \( X = C(E) \) instead, and closedness of \( C(E) \) in \( B(E) \) is the uniform limit theorem. The converse implication (ii) \( \Rightarrow \) (i) explains why no version of the M-test can survive in an incomplete space, and the same skeleton — dominate, sum the dominants, invoke completeness — reappears in Bochner integration, in the Weierstrass factorisation of entire functions via uniformly convergent products, and in the convergence theory of Neumann series \( \sum T^n \) in operator algebras.

Common misconceptions. The M-test does not characterise uniform convergence (it is one-directional); it proves absolute convergence as a by-product, so it can never establish uniform convergence of a conditionally convergent series — for those one needs Dirichlet's or Abel's uniform tests; and it never, by itself, licenses term-by-term differentiation.

Worked examples

Example 1. Show that \( f(x) = \sum_{n=1}^{\infty} \frac{\cos(nx)}{n^2} \) defines a continuous function on all of \( \mathbb{R} \).

Example 2. Show that \( g(x) = \sum_{n=1}^{\infty} n e^{-nx} \) converges uniformly on \( [\delta, \infty) \) for every \( \delta \gt 0 \), hence defines a continuous function on \( (0, \infty) \) — but that the convergence is not uniform on \( (0, \infty) \).

Problems

1. Prove that \( \sum_{n=1}^{\infty} \frac{\sin(nx)}{n^3} \) converges uniformly on \( \mathbb{R} \) and that its sum is continuous.
   Solution

   For all \( x \in \mathbb{R} \), \( \left| \frac{\sin(nx)}{n^3} \right| \le \frac{1}{n^3} =: M_n \), since \( |\sin t| \le 1 \). The series \( \sum \frac{1}{n^3} \) is a convergent \( p \)-series (\( p = 3 \gt 1 \)). By the Weierstrass M-test the function series converges uniformly on \( \mathbb{R} \). Each term \( x \mapsto \frac{\sin(nx)}{n^3} \) is continuous, so every partial sum is continuous, and by the uniform limit theorem the sum is continuous on \( \mathbb{R} \). \( \blacksquare \)
2. Let \( f_n(x) = x^n (1 - x) \) on \( [0,1] \). Compute \( \sup_{x \in [0,1]} |f_n(x)| \), show the M-test does not apply, and decide (with proof) whether \( \sum_{n=0}^{\infty} f_n \) converges uniformly on \( [0,1] \).
   Solution

   On \( [0,1] \), \( f_n \ge 0 \) and \( f_n'(x) = x^{n-1} \left( n - (n+1)x \right) \), so the maximum is at \( x_n = \frac{n}{n+1} \): \[ \sup_{[0,1]} f_n = \left( \frac{n}{n+1} \right)^n \frac{1}{n+1} = \frac{1}{n+1} \left( 1 + \frac{1}{n} \right)^{-n} \to 0 \cdot e^{-1}, \quad \text{i.e. } \sup f_n \sim \frac{1}{e\,n} . \] Any admissible \( M_n \) must be at least this supremum, and \( \sum \frac{1}{e\,n} \) diverges (harmonic), so the M-test cannot apply. Directly: the partial sums telescope for \( x \in [0,1) \): \[ S_N(x) = (1-x) \sum_{n=0}^{N} x^n = 1 - x^{N+1} \to 1, \] while \( S_N(1) = 0 \to 0 \). The pointwise sum is \( 1 \) on \( [0,1) \) and \( 0 \) at \( x = 1 \): discontinuous. Since every \( S_N \) is continuous, a uniform limit would be continuous (uniform limit theorem); hence the convergence is not uniform on \( [0,1] \). (It is uniform on each \( [0, r] \), \( r \lt 1 \), since there \( \sup |1 - S_N| = r^{N+1} \to 0 \).) \( \blacksquare \)
3. Let \( \sum_{n=0}^{\infty} a_n x^n \) be a power series with radius of convergence \( R \gt 0 \). Prove that it converges uniformly on \( [-r, r] \) for every \( 0 \lt r \lt R \), and deduce that its sum is continuous on \( (-R, R) \).
   Solution

   Fix \( 0 \lt r \lt R \). Since \( r \lt R \), the point \( x = r \) lies strictly inside the interval of convergence, and by the Cauchy–Hadamard theorem (root-test form: \( \limsup_n |a_n|^{1/n} = \frac{1}{R} \)) the series converges absolutely at \( x = r \); that is, \( \sum_{n=0}^{\infty} |a_n| r^n \lt \infty \). For \( |x| \le r \), \[ |a_n x^n| \le |a_n| r^n =: M_n , \] a bound independent of \( x \), with \( \sum M_n \lt \infty \). The Weierstrass M-test gives uniform convergence on \( [-r, r] \). Each monomial \( a_n x^n \) is continuous, so the sum is continuous on \( [-r, r] \) by the uniform limit theorem. Every point of \( (-R, R) \) lies in the interior of some \( [-r, r] \) with \( r \lt R \), and continuity is local, so the sum is continuous on \( (-R, R) \). \( \blacksquare \)
4. Define \( W(x) = \sum_{n=0}^{\infty} 2^{-n} \cos(4^n x) \). (a) Show \( W \) is continuous on \( \mathbb{R} \). (b) Show that the formally differentiated series \( -\sum_{n=0}^{\infty} 2^n \sin(4^n x) \) does not converge uniformly on \( \mathbb{R} \), so the term-by-term differentiation theorem cannot be applied to \( W \).
   Solution

   (a) \( |2^{-n} \cos(4^n x)| \le 2^{-n} =: M_n \) for all \( x \), and \( \sum 2^{-n} = 2 \lt \infty \) (geometric, ratio \( \frac{1}{2} \)). By the M-test the series converges uniformly on \( \mathbb{R} \); the terms are continuous, so \( W \) is continuous by the uniform limit theorem. (b) If a series \( \sum g_n \) converges uniformly on \( E \), its terms satisfy \( g_n = S_n - S_{n-1} \to 0 \) uniformly, i.e. \( \sup_E |g_n| \to 0 \). Here \( g_n(x) = -2^n \sin(4^n x) \) and \[ \sup_{x \in \mathbb{R}} |g_n(x)| = 2^n \quad \left( \text{attained at } x = \frac{\pi}{2 \cdot 4^n} \right) , \] which tends to \( \infty \), not \( 0 \). Hence the derived series does not converge uniformly on \( \mathbb{R} \) (indeed not even pointwise at typical points), and the hypothesis of the term-by-term differentiation theorem — uniform convergence of the derived series — fails. (Weierstrass proved more: for suitable parameters such functions are differentiable at no point whatsoever; that requires a separate, harder argument and does not follow from the M-test alone.) \( \blacksquare \)
5. (Converse fails.) Let \( f_n(x) = \frac{(-1)^n}{n + x} \) on \( E = [0, \infty) \). Prove: (a) \( \sum_{n=1}^{\infty} f_n \) converges uniformly on \( E \); (b) the convergence is not absolute at any point of \( E \); (c) there is no sequence \( (M_n) \) with \( |f_n(x)| \le M_n \) on \( E \) and \( \sum M_n \lt \infty \). Conclude that the Weierstrass M-test admits no converse.
   Solution

   (a) Fix \( x \ge 0 \). The numbers \( a_n(x) = \frac{1}{n+x} \) are positive, strictly decreasing in \( n \), and tend to \( 0 \), so by the alternating series test \( \sum (-1)^n a_n(x) \) converges, and by the alternating-series remainder estimate, \[ \left| \sum_{n=N+1}^{\infty} \frac{(-1)^n}{n+x} \right| \le a_{N+1}(x) = \frac{1}{N+1+x} \le \frac{1}{N+1} . \] The bound \( \frac{1}{N+1} \) is independent of \( x \) and tends to \( 0 \), so \( \sup_{x \ge 0} |f(x) - S_N(x)| \le \frac{1}{N+1} \to 0 \): the convergence is uniform on \( [0, \infty) \). (b) For fixed \( x \ge 0 \), \( \sum_{n=1}^{\infty} |f_n(x)| = \sum_{n=1}^{\infty} \frac{1}{n+x} \) diverges by limit comparison with the harmonic series ( \( \frac{1/(n+x)}{1/n} \to 1 \) ). So the convergence is conditional at every point. (c) Suppose \( |f_n(x)| \le M_n \) for all \( x \ge 0 \). Taking \( x = 0 \) gives \( M_n \ge \frac{1}{n} \), whence \( \sum M_n \ge \sum \frac{1}{n} = \infty \). No summable dominating sequence exists. Conclusion: uniform convergence on \( E \) neither implies absolute convergence nor the existence of summable dominants; the implication in the M-test is strictly one-way. (Positive counterpart: for alternating-type series, Dirichlet's uniform test and Abel's uniform test are the correct tools.) \( \blacksquare \)