Answer to Question #170869 in Real Analysis for Prathibha Rose

Question

State and prove weierstrass M- test

Answer 1

Weierstrass M-test.

 Suppose that (f_n) is a sequence of real- or complex-valued functions defined on a set A, and that

there is a sequence of non-negative numbers (M_n) satisfying

${\displaystyle \forall n\geq 1,\forall x\in A:\ |f_{n}(x)|\leq M_{n},}\\ {\displaystyle \{\displaystyle \sum _{n=1}^{\infty }M_{n}<\infty .}\\ \text{Then the series}\\ {\displaystyle \sum _{n=1}^{\infty }f_{n}(x)}\\ \text{ converges absolutely and uniformly on A}$

Proof

$\text{Consider the sequence of functions}\\ {\displaystyle S_{n}(x)=\sum _{k=1}^{n}f_{k}(x).}\\ \text{Since the series} {\displaystyle \sum _{n=1}^{\infty }M_{n}} converges and Mn ≥ 0 \text{for every} n, \text{then by the Cauchy criterion},\\ {\displaystyle \forall \varepsilon >0:\exists N:\forall m>n>N:\sum _{k=n+1}^{m}M_{k}<\varepsilon .}\\ \text{For the chosen N},\\ {\displaystyle \forall x\in A:\forall m>n>N}\\ {\displaystyle \left|S_{m}(x)-S_{n}(x)\right|=\left|\sum _{k=n+1}^{m}f_{k}(x)\right|{\overset {(1)}{\leq }}\sum _{k=n+1}^{m}|f_{k}(x)|\leq \sum _{k=n+1}^{m}M_{k}<\varepsilon .}\\ \text{(Inequality (1) follows from the triangle inequality.)}\\ \text{The sequence Sn(x) is thus a Cauchy sequence in R or C, and by completeness, it converges to some number S(x) that depends on x. For n > N we can write}\\ {\displaystyle \left|S(x)-S_{n}(x)\right|=\left|\lim _{m\to \infty }S_{m}(x)-S_{n}(x)\right|=\lim _{m\to \infty }\left|S_{m}(x)-S_{n}(x)\right|\leq \varepsilon .}\\ \text{Since N does not depend on x, this means that the sequence Sn of partial sums converges uniformly to the function S}. \text{Hence, by definition, the series} {\displaystyle \sum _{k=1}^{\infty }f_{k}(x)} \text{converges uniformly}.\\ \text{Analogously, one can prove that} {\displaystyle \sum _{k=1}^{\infty }|f_{k}(x)|} \text{converges uniformly.}$