Proof of The Weierstrass M-Test

Each result cited below is universally quantified over the data in its own statement, and is applied to the data named in the step where it is cited. Write $\Sigma=\sum_{k=1}^{\infty}M_{k}$ and, for $m\in\mathbb{N}$, $\sigma_{m}=\sum_{k=1}^{m}M_{k}$, so that $\tau_{m}=\Sigma-\sigma_{m}$.

Claim 1. 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. The domination clause An Absolutely Convergent Series of Real Numbers Converges (iii), 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; An Absolutely Convergent Series of Real Numbers Converges (i) then shows that it converges.

Claim 2.

Step 1 (the numbers $\tau_{m}$). Since $0\le M_{k}$ for every $k\in\mathbb{N}$ and $\sum_{k=1}^{\infty}M_{k}$ converges, the domination clause Series of Nonnegative Real Numbers, Comparison, and the Geometric Series (ii) gives

$$0\le\sigma_{m}\le\Sigma\qquad\text{for every }m\in\mathbb{N}.$$

By claim 3 of Elementary Arithmetic in an Ordered Field, the inequality $\sigma_{m}\le\Sigma$ is exactly $0\le\Sigma-\sigma_{m}=\tau_{m}$.

The sequence $(\sigma_{m})_{m\in\mathbb{N}}$ converges to $\Sigma$ by Series of Real Numbers (ii). The constant sequence with every term equal to $\Sigma$ converges to $\Sigma$ directly from Limit of a Sequence of Real Numbers, since $|\Sigma-\Sigma|=0<\varepsilon$ for every real $\varepsilon>0$. By claim 3 of Arithmetic of Limits of Real Sequences, applied with the convergent sequence $(\sigma_{m})_{m\in\mathbb{N}}$ in both sequence slots and with the scalar $-1$, the sequence $(-\sigma_{m})_{m\in\mathbb{N}}$ converges to $-\Sigma$; by claim 1 of that theorem, applied to the constant sequence and to $(-\sigma_{m})_{m\in\mathbb{N}}$, the sequence $(\tau_{m})_{m\in\mathbb{N}}$ converges to $\Sigma+(-\Sigma)=0$.

Step 2 (a block estimate). Assume now that $f:D\to\mathbb{R}$ satisfies $f(x)=\sum_{k=1}^{\infty}g_{k}(x)$ for every $x\in S$, and fix $x\in S$ and $m\in\mathbb{N}$. Let $p\in\mathbb{N}$.

Apply Splitting a Finite Sum at an Index with the field $\mathbb{R}$, the natural numbers $m$ and $p$, and the map $k\mapsto g_{k}(x)$ on the initial segment $[m+p]$. Writing $u_{j}=g_{m+j}(x)$ for $j\in[p]$, and recalling from Series of Real-Valued Functions and Their Partial Sums (i) that $s_{n}(x)=\sum_{k=1}^{n}g_{k}(x)$, that lemma gives

$$s_{m+p}(x)=s_{m}(x)+\sum_{j=1}^{p}u_{j}.$$

The same lemma applied to the map $k\mapsto M_{k}$ on $[m+p]$ gives $\sigma_{m+p}=\sigma_{m}+\sum_{j=1}^{p}M_{m+j}$.

By claim 2 of Comparison and Absolute Value Bounds for Finite Sums of Real Numbers, and then by claim 1 of that lemma applied to the bounds $|u_{j}|=|g_{m+j}(x)|\le M_{m+j}$, which hold because $x\in S$,

$$\Bigl|\sum_{j=1}^{p}u_{j}\Bigr|\le\sum_{j=1}^{p}|u_{j}|\le\sum_{j=1}^{p}M_{m+j}=\sigma_{m+p}-\sigma_{m}.$$

Step 1 gives $\sigma_{m+p}\le\Sigma$, so $\bigl(\Sigma-\sigma_{m}\bigr)-\bigl(\sigma_{m+p}-\sigma_{m}\bigr)=\Sigma-\sigma_{m+p}$ is nonnegative by claim 3 of Elementary Arithmetic in an Ordered Field, and a second use of that claim gives $\sigma_{m+p}-\sigma_{m}\le\Sigma-\sigma_{m}=\tau_{m}$. Since $s_{m+p}(x)-s_{m}(x)=\sum_{j=1}^{p}u_{j}$, we conclude

$$|s_{m+p}(x)-s_{m}(x)|\le\tau_{m}\qquad\text{for every }p\in\mathbb{N}.$$

Step 3 (passing to the limit in $p$). By Claim 1 and Series of Real Numbers (ii), the sequence $(s_{n}(x))_{n\in\mathbb{N}}$ converges to $\sum_{k=1}^{\infty}g_{k}(x)=f(x)$. The sequence $(s_{m+p}(x))_{p\in\mathbb{N}}$ also converges to $f(x)$: given a real $\varepsilon>0$, choose $N\in\mathbb{N}$ with $|s_{n}(x)-f(x)|<\varepsilon$ for every $n\in\mathbb{N}$ with $N\le n$; by claim 6 of Properties of the Order on the Natural Numbers we have $p<p+m$, hence $p\le p+m$ by claim 1 of that lemma, and $p+m=m+p$ by claim 4 of Arithmetic of Addition on the Natural Numbers; so if $N\le p$ then $N\le m+p$, by the transitivity of $\le$ recorded in claim 1 of Properties of the Order on the Natural Numbers and hence $|s_{m+p}(x)-f(x)|<\varepsilon$.

The constant sequence with every term equal to $s_{m}(x)$ converges to $s_{m}(x)$, as in Step 1. Hence, by claim 3 of Arithmetic of Limits of Real Sequences applied with the constant sequence in both sequence slots and with the scalar $-1$, and then by claim 1 of that theorem, applied to $\bigl(s_{m+p}(x)\bigr)_{p\in\mathbb{N}}$ and to the constant sequence with every term equal to $-s_{m}(x)$, the sequence $\bigl(s_{m+p}(x)-s_{m}(x)\bigr)_{p\in\mathbb{N}}$ converges to $f(x)-s_{m}(x)$; by claim 4 of Order Properties of Limits of Real Sequences the sequence $\bigl(|s_{m+p}(x)-s_{m}(x)|\bigr)_{p\in\mathbb{N}}$ converges to $|f(x)-s_{m}(x)|$. The constant sequence with every term equal to $\tau_{m}$ converges to $\tau_{m}$, and by Step 2 each term of the former sequence is at most the corresponding term of the latter, so claim 1 of Order Properties of Limits of Real Sequences gives

$$|f(x)-s_{m}(x)|\le\tau_{m} .$$

As $x\in S$ and $m\in\mathbb{N}$ were arbitrary, this is the asserted bound.

Claim 3. Let $\varepsilon$ be a real number with $0<\varepsilon$. By Claim 2 the sequence $(\tau_{m})_{m\in\mathbb{N}}$ converges to $0$, so Limit of a Sequence of Real Numbers provides $K\in\mathbb{N}$ such that $|\tau_{m}-0|<\varepsilon$ for every $m\in\mathbb{N}$ with $K\le m$. Since $0\le\tau_{m}$, Absolute Value in an Ordered Field gives $|\tau_{m}|=\tau_{m}$, so $\tau_{m}<\varepsilon$ for every such $m$.

Let $m\in\mathbb{N}$ with $K\le m$ and let $x\in S$. By claim 2 of Properties of the Absolute Value in an Ordered Field and Claim 2 above,

$$|s_{m}(x)-f(x)|=|f(x)-s_{m}(x)|\le\tau_{m}<\varepsilon .$$

This is exactly the condition of Pointwise and Uniform Convergence of a Sequence of Real-Valued Maps (ii) for the sequence of partial sums $(s_{m})_{m\in\mathbb{N}}$ and the function $f$ on $S$, so that sequence converges uniformly to $f$ on $S$; by Series of Real-Valued Functions and Their Partial Sums (iii) the series $\sum_{k=1}^{\infty}g_{k}$ converges uniformly to $f$ on $S$.