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.

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 $x$.

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

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}$$\mathbb{R}$, take $f_n(x)=2^{-n}\cos(3^nx)$$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 $\mathbb{R}$:

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¶

• Uniform Norm
• Infinite Series of Functions

Related Concepts¶

• Normal Convergence
• Weierstrass Function

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