Proof of The Weierstrass M-Test

By Sequences of Real-Valued Maps: the Map of Pointwise Limits on a Subset, and the Sequence of Pointwise Partial Sums (ii), applied to $(g_{k})_{k\in\mathbb{N}}$, there is exactly one sequence $(s_{n})_{n\in\mathbb{N}}$ in $\mathbb{R}^{D}$ with $s_{n}(x)=\sum_{k=1}^{n}g_{k}(x)$ for every $n\in\mathbb{N}$ and every $x\in D$, so its terms are the maps $s_{n}$ of the statement; and for every $x\in S$, $(s_{n}(x))_{n\in\mathbb{N}}$ is the sequence of partial sums of $(g_{k}(x))_{k\in\mathbb{N}}$.

Proof of (i). Fix $x\in S$. By hypothesis $|g_{k}(x)|\le M_{k}$ for every $k\in\mathbb{N}$, and $\sum_{k=1}^{\infty}M_{k}$ converges; so Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (iv), applied to the sequences $(g_{k}(x))_{k\in\mathbb{N}}$ and $(M_{k})_{k\in\mathbb{N}}$, shows that $\sum_{k=1}^{\infty}g_{k}(x)$ converges absolutely, and Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (iii) shows that it converges. By Series of Real Numbers: Partial Sums, Convergence, the Sum and Absolute Convergence (ii), its sequence of partial sums $(s_{n}(x))_{n\in\mathbb{N}}$ is convergent, and $\sum_{k=1}^{\infty}g_{k}(x)$ is its limit $\lim_{n\to\infty}s_{n}(x)$.

So $(s_{n}(x))_{n\in\mathbb{N}}$ is convergent for every $x\in S$, and Sequences of Real-Valued Maps: the Map of Pointwise Limits on a Subset, and the Sequence of Pointwise Partial Sums (i), applied to the set $D$, the subset $S$ and the sequence $(s_{n})_{n\in\mathbb{N}}$ in $\mathbb{R}^{D}$, gives exactly one map $s:S\to\mathbb{R}$ with $s(x)=\lim_{n\to\infty}s_{n}(x)=\sum_{k=1}^{\infty}g_{k}(x)$ for every $x\in S$.

Proof of (ii). Fix $n\in\mathbb{N}$. Since $\sum_{k=1}^{\infty}M_{k}$ converges, Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping (iv), applied to $(M_{k})_{k\in\mathbb{N}}$ with $p=n$, shows that $\sum_{k=1}^{\infty}M_{n+k}$ converges.

Fix $x\in S$. By (i), $\sum_{k=1}^{\infty}g_{k}(x)$ converges, so Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping (iv), applied to $(g_{k}(x))_{k\in\mathbb{N}}$ with $p=n$, shows that $\sum_{k=1}^{\infty}g_{n+k}(x)$ converges and

$$s(x)=\sum_{k=1}^{\infty}g_{k}(x)=\sum_{k=1}^{n}g_{k}(x)+\sum_{k=1}^{\infty}g_{n+k}(x)=s_{n}(x)+\sum_{k=1}^{\infty}g_{n+k}(x),$$

that is, $s(x)-s_{n}(x)=\sum_{k=1}^{\infty}g_{n+k}(x)$. For every $k\in\mathbb{N}$, $n+k\in\mathbb{N}$ by Natural Numbers Are the Successors in Omega: One Is Least and Not a Successor of a Natural Number, the Successor Is Injective, and N Is Closed under Addition and Multiplication (vi), so the hypothesis gives $|g_{n+k}(x)|\le M_{n+k}$. Hence Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (iv), applied to the sequences $(g_{n+k}(x))_{k\in\mathbb{N}}$ and $(M_{n+k})_{k\in\mathbb{N}}$, gives

$$|s(x)-s_{n}(x)|=\Big|\sum_{k=1}^{\infty}g_{n+k}(x)\Big|\le\sum_{k=1}^{\infty}M_{n+k}.$$

By Elementary Properties of Series of Real Numbers: Linearity, Null Terms, the Cauchy Criterion, Index Shifts, Tails, Order and Telescoping (v), applied to $(M_{k})_{k\in\mathbb{N}}$, the sequence $\big(\sum_{k=1}^{\infty}M_{n+k}\big)_{n\in\mathbb{N}}$ is a null sequence, that is, it converges to $0$ in the sense of Convergent Sequences of Real Numbers (i). If $M_{k}\ge0$ for every $k\in\mathbb{N}$, then $0\le M_{k}\le M_{k}$ for every $k\in\mathbb{N}$, and Series of Real Numbers with Nonnegative Terms, Comparison, Absolute Convergence and the Geometric Series (ii), applied with $(M_{k})_{k\in\mathbb{N}}$ in place of both $(a_{k})$ and $(b_{k})$, gives $0\le\sum_{k=1}^{\infty}M_{n+k}$ for every $n\in\mathbb{N}$.

Proof of (iii). By (ii), the sequence $\big(\sum_{k=1}^{\infty}M_{q+k}\big)_{q\in\mathbb{N}}$ converges to $0$. The choices are made in this order: a real $\varepsilon>0$ is given; then $K\in\mathbb{N}$ is taken, by Convergent Sequences of Real Numbers (i) with limit $0$, such that $\big|\sum_{k=1}^{\infty}M_{q+k}\big|<\varepsilon$ for every $q\in\mathbb{N}$ with $q\ge K$; then $n\in\mathbb{N}$ with $n\ge K$ and $x\in S$ are arbitrary. By Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (ix), $\sum_{k=1}^{\infty}M_{n+k}<\varepsilon$. By (ii) and Inequalities in an Ordered Field: Mixed Transitivity, Strict Sums, Signs, Products, Natural Numbers, Halving, Reciprocals, Absolute Values and Squares (i),

$$|s_{n}(x)-s(x)|=|s(x)-s_{n}(x)|\le\sum_{k=1}^{\infty}M_{n+k}<\varepsilon .$$

This is the condition of Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii), read with the sequence $(s_{n})_{n\in\mathbb{N}}$ in $\mathbb{R}^{D}$, $E=S$ and $s$ in place of $f$; so $(s_{n})_{n\in\mathbb{N}}$ converges uniformly on $S$ to $s$. Finally, let $f:E\to\mathbb{R}$ with $S\subseteq E$ and $f(x)=\sum_{k=1}^{\infty}g_{k}(x)$ for every $x\in S$. Then $f(x)=s(x)$ for every $x\in S$ by (i), and the condition of Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii) involves the limit map only through its values at points of $S$; so the same $K$ serves for $f$, and $(s_{n})_{n\in\mathbb{N}}$ converges uniformly on $S$ to $f$.