Weierstrass $M$ -test¶

Theorem¶

Suppose $(f_n)_{n=1}^\infty$ is a sequence of bounded functions on $X$ such that the series $\displaystyle \sum_{n=1}^\infty \|f_n\|_\infty$ converges. Then, the series $\displaystyle \sum_{n=1}^\infty f_n(x)\rightrightarrows f\text{.}$

The Uniform Norm $\|f_n\|_\infty$ is the smallest common upper bound on $|f_n(x)|$ over all $x$ . Summability of these bounds is what makes the tail estimate independent of the input.

Interpretation¶

The $M$ -test reduces uniform convergence of a series of functions to convergence of one numerical series. Bound each term by a constant $M_n$ that is valid at every $x\in X$ $X$ ; if $\displaystyle \sum_n M_n$ converges, the series of functions converges uniformly. The smallest such constant is $\|f_n\|_\infty$ , so the theorem is stated with the uniform norms, and any larger summable bounds $M_n$ work by comparison.

Three panels with functions f1, f2 and f3 oscillating inside shaded bands of half-width M1, M2 and M3, which shrink from panel to panel.

Each panel shows one term $f_n$ inside a band of half-width $M_n$ , and the bands shrink from panel to panel. The band bounds the whole graph of $f_n$ , not just its value at one point, which is what makes the estimate below independent of $x$ .

Here $|f_n(x)|\le M_n=1/n^2$ for every $x$ . The tail bound $\displaystyle \sum_{n>N}M_n$ is one band width for all $x$ at once, which gives uniform convergence.

For partial sums $\displaystyle S_N=\sum_{n=1}^N f_n$ and $K>N$ $N$ , the same bounds control every tail:

|S_K(x)-S_N(x)|\le\sum_{n=N+1}^{K}M_n.

The right-hand side tends to zero independently of $x$ . For each fixed $x$ , the partial sums therefore form a Cauchy sequence of real numbers, so by the Cauchy criterion they converge, which gives a pointwise limit $f$ . Letting $K\to\infty$ gives $\displaystyle |f(x)-S_N(x)|\le\sum_{n>N}M_n$ for all $x$ , so the convergence is uniform.

The hypothesis is normal convergence. The test is the standard tool for proving that a power series converges uniformly on closed intervals inside its radius, and that the Weierstrass function is continuous.

A common slip is to use a bound that depends on $x$ . On $(-1,1)$ , $|x^n|\le|x|^n$ with $\displaystyle \sum_n|x|^n<\infty$ for each fixed $x$ , but this gives only pointwise convergence, and the geometric series is indeed not uniformly convergent there. The test is also sufficient, not necessary: the constant functions $(-1)^n/n$ form a uniformly convergent series, although $\displaystyle \sum_n 1/n$ diverges.

Key Results¶

The limit function $f$ is also bounded on $X$ .
Due to the uniform limit theorem, if each $f_n$ is continuous, then the limit function $f$ is continuous.

Examples¶

On $X=\mathbb{R}$ $R$ , take $f_n(x)=2^{-n}\cos(3^nx)$ $2^{- n} cos (3^{n} x)$ . Since $|\cos(3^nx)|\le1$ for every real $x$ , the bound $|f_n(x)|\le M_n=2^{-n}$ holds simultaneously for all $x$ . The numerical series of bounds sums to $1$ , so the function series converges uniformly on $\mathbb{R}$ :

\sup_{x\in\mathbb{R}}|f(x)-S_N(x)|\le\sum_{n=N+1}^\infty 2^{-n}=2^{-N}.

For example, $N=10$ makes the error at most $2^{-10}=1/1024$ at every real $x$ , using the same number of terms everywhere.

Prerequisites¶

Sources¶

Math III: Analysis · Lecture 2

Aliases: M-test, Weierstrass test, m test Symbols: element of, infinity, real numbers, sum

Last updated: 20 October 2026 03:02

Weierstrass MMM-test¶