# Weierstrass M-test

In mathematics, the Weierstrass M-test is a test for determining whether an infinite series of functions converges uniformly and absolutely. It applies to series whose terms are functions with real or complex values, and is analogous to the comparison test for determining the convergence of series of real or complex numbers.

## Statement

Weierstrass M-test. Suppose that {fn} is a sequence of real- or complex-valued functions defined on a set A, and that there is a sequence of positive numbers {Mn} satisfying

$\forall n\geq 1,\forall x\in A:\ |f_{n}(x)|\leq M_{n},$ $\sum _{n=1}^{\infty }M_{n}<\infty$ Then the series

$\sum _{n=1}^{\infty }f_{n}(x)$ converges absolutely and uniformly on A.

Remark. The result is often used in combination with the uniform limit theorem. Together they say that if, in addition to the above conditions, the set A is a topological space and the functions fn are continuous on A, then the series converges to a continuous function.

## Generalization

A more general version of the Weierstrass M-test holds if the codomain of the functions {fn} is any Banach space, in which case the statement

$|f_{n}|\leq M_{n}$ may be replaced by

$\|f_{n}\|\leq M_{n}$ ,

where $\|\cdot \|$ is the norm on the Banach space. For an example of the use of this test on a Banach space, see the article Fréchet derivative.

## Proof

Consider the sequence of functions

$S_{n}(x)=\sum _{k=1}^{n}f_{k}(x)$ Since the series $\sum _{n=1}^{\infty }M_{n}$ converges and Mn ≥ 0 for every n, then by the Cauchy criterion,

$\forall \varepsilon >0:\exists N:\forall n>m>N:\sum _{k=m+1}^{n}M_{k}<\varepsilon .$ For the chosen N,

$\forall x\in A:\forall n>m>N$ $\left|S_{n}(x)-S_{m}(x)\right|=\left|\sum _{k=m+1}^{n}f_{k}(x)\right|{\overset {(1)}{\leq }}\sum _{k=m+1}^{n}|f_{k}(x)|\leq \sum _{k=m+1}^{n}M_{k}<\varepsilon .$ Thus the sequence of partial sums of the series converges uniformly. Hence, by definition, the series $\sum _{k=1}^{\infty }f_{k}(x)$ converges uniformly.

Inequality (1) follows from the triangle inequality.